Contentstomasz.przebinda.com/LieTheoryII_Spring_2020.pdf · Problems arising in Physics and Number Theory motivated a rapid growth of Harmonic Analysis on non-commutative groups

LIE THEORY II, SPRING 2020

TOMASZ PRZEBINDA

Contents

1. Introduction 22. The Fourier Transform on the Schwartz space S(R) 33. Magnetic Resonance Imaging 74. The Fourier Transform on the space C∞(U1) 85. Symmetries of platonic solids 106. Basic representation theory of finite groups 147. Some Functional Analysis on a Hilbert space 268. Representations of locally compact groups 289. Haar measures and extension of a representation of G to a

representation of L1(G) 3010. Representations of a compact group 3110.1. Unique factorization domains and Hilbert’s Nulstellensatz 3810.2. Spherical harmonics in Rn 4111. Square integrable representations of a locally compact group 4512. G = SL2(R) 4712.1. The finite dimensional unitary representations 4712.2. The maximal compact subgroup K = SO2(R) 4912.3. Induced representations 5212.4. Finite dimensional representations 5412.5. Smooth vectors and analytic vectors 5512.6. The derivative of the right regular representation 5612.7. The universal enveloping algebra 5712.8. K-multiplicity 1 representations 5712.9. The character of a (g,K)-module 6012.10. The unitary dual 6112.11. The character of the sum of discrete series 6412.12. Harish-Chandra’s Plancherel formula for L2(G) 6512.13. An irreducible unitary non-tempered representation of SL2(R). 7113. The Heisenberg group. 7313.1. Structure of the group. 7313.2. Decomposition of the right regular representation 7513.3. Structure of the Lie algebra. 78

Date: May 4, 2020.

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13.4. A generic representation of the Lie algebra. 7814. An example of a wild group. 7814.1. The Mautner group 7914.2. The right regular representation 1 7914.3. The right regular representation 2 8214.4. A surprise 84References 85

1. Introduction

Classical Harmonic Analysis concerns a decomposition of a function (signal) into asuperposition of components corresponding to simple harmonics. The analysis of thesignal aims at finding these components and the synthesis is the reconstruction of thesignal out of them. There is often some “signal processing”, dictated by applications,between the analysis and synthesis.

The simple harmonics behave well under various symmetries and this is the reasonfor the decomposition. The fundamental results are Parseval’s Theorem (1806) for theFourier series, [Par06], and Plancherel’s Theorem (1910) for the Fourier transform, [Pla10].Among the best known applications is the Magnetic Resonance Imaging, for which PeterMansfield and Paul Lauterbur were awarded a Nobel prize in 2003.

If the function is defined on a commutative group, such as the additive group of the realnumbers or the multiplicative group of the complex numbers of absolute value one, thenthe simple harmonics are the eigenvectors under the translations. This is the ultimatesymmetry one could expect.

Problems arising in Physics and Number Theory motivated a rapid growth of HarmonicAnalysis on non-commutative groups. The earliest examples were the Heisenberg group,necessary for a formulation of the principles of Quantum Mechanics (J. Von Neumann1926, [vN26]), and the compact Lie groups (Peter-Weyl 1927, [PW27]). Here the simpleharmonics are replaced by irreducible unitary representations. All of them may be found byanalyzing the square integrable functions on these groups, so that an analog of Plancherel’sTheorem may be viewed as the top achievement of the theory.

However, there are plenty of other groups of interest which have irreducible unitaryrepresentations occurring outside the space of the square integrable functions on thegroup. The main class are the non-compact semisimple Lie groups, such as SL2(R).Though the irreducible unitary representations of most of them are not understood yet,the representations that can be found in the space of the square integrable functions onthe group are known and the decomposition of an arbitrary such function in terms of theserepresentations is known as the Plancherel formula. For the group SLn(C) this formulawas first found by Gelfand and Naimark in 1950, , and for SLn(R) by Gelfand and Graevin 1953, . The Plancherel formula on an arbitrary Real Reductive Group was publishedby Harish-Chandra in 1976, [Har76], and is considered as one of the greatest achievementsof Mathematics of the 20th century.

LIE THEORY II, SPRING 2020 3

A goal of these lectures is to explain the ingredients of Harish-Chandra’s Plancherelformula, explain how they fit together, study particular cases and go through all thedetails for the group of the real unimodular matrices of size two.

All the necessary information in its original nearly perfect form is contained in Harish-Chadra’s articles [HC14a], [HC14b], [HC14d], [HC14c], [HC18]. The example SL2(R) isexplained in classical books such as [Lan75]. For all of that, a good understanding ofthe Fourier Transform and Distribution theory on an Euclidean space is needed. HereHormender’s “The Analysis of Linear Partial Differential Operators I” is one of the bestreferences, [Hor83].

2. The Fourier Transform on the Schwartz space S(R)

Here we follow [Hor83, section 7.1]. Recall that the Schwartz space S(R) consists of allinfinitely many times differentiable functions f : R → C such that for any two integersn, k ≥ 0

supx∈R|xn∂kxf(x)| <∞ .

In particular S(R) ⊆ L1(R) and we have the well defined

Theorem 1. The Fourier transform

f(y) = Ff(y) =

∫Re−2πixyf(x) dx (y ∈ R, f ∈ S(R)) (1)

maps the Schwartz space S(R) into itself, is invertible, and the inverse is given by

f(x) =

∫Re2πixyFf(y) dy (x ∈ R, f ∈ S(R)) . (2)

f(0) =

∫RFf(y) dy (f ∈ S(R)) .

Sinced

dyFf(y) =

∫Re−2πixy(−2πix)f(x) dx (3)

and ∫Re−2πixyf ′(x) dx = 2πiyFf(y) , (4)

the inclusion FS(R) ⊆ S(R) is easy to check. For the rest we need a few lemmas.

Lemma 2. Let f ∈ S(R) be such that f(0) = 0. Set g(x) =∫ 1

0f ′(tx) dt. Then f(x) =

xg(x) and g ∈ S(R).

Proof. The equality f(x) = xg(x) is immediate from the Fundamental Theorem of Cal-culus, via a change of variables y = tx.

Fix two non-negative integers n and k. Suppose |x| ≤ 1. Then∣∣xng(k)(x)∣∣ =

∣∣∣∣xn ∫ 1

0

tkf (k+1)(tx) dt

∣∣∣∣ ≤ ∫ 1

0

|f (k+1)(tx)| dt ≤ maxy∈R|f(y)| <∞.

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Notice that

xng(k)(x) = xn(d

dx

)k(x−1f(x))

= xnk∑p=0

k!

p!(k − p)!(−1)(−2) . . . (−p)x−p−1f (k−p)(x)

and that

max|x|≥1

∣∣xn−p−1f (k−p)(x)∣∣ <∞.

Hence

maxx∈R|xng(k)(x)| <∞.

�

Corollary 3. Fix y ∈ R. Let φ ∈ S(R) be such that φ(y) = 0. Then that there is ψ ∈ S(R)such that

φ(x) = (x− y)ψ(x).

Lemma 4. Let T : S(R) → S(R) be a linear map with the property that if φ(y) = 0 forsome y ∈ R then Tφ(y) = 0 for the same y. Then there is a function c(x) such that

Tφ(x) = c(x)φ(x) (φ ∈ S(R).

(In other words, T is the multiplication by the function c.)

Proof. Let φ1(x) = e−x2. As we know this function belongs to S(R). Fix x ∈ R. Then

for any φ2 ∈ S(R)

(φ2(x)φ1 − φ1(x)φ2) (x) = φ2(x)φ1(x)− φ1(x)φ2(x) = 0.

Hence, by the assumption on T ,

0 = T (φ2(x)φ1 − φ1(x)φ2) (x). (5)

Since T is linear

T (φ2(x)φ1) = φ2(x)T (φ1) and T (φ1(x)φ2) = φ1(x)T (φ2).

Thus evaluation at x and using (5) we see that

0 = φ2(x)T (φ1)(x)− φ1(x)T (φ2)(x).

Therefore

T (φ2)(x) =T (φ1)(x)

φ1(x)φ2(x).

Thus the claim holds with

c(x) =T (φ1)(x)

φ1(x).

�


Lemma 5. Suppose T : S(R)→ S(R) is a linear map which commutes with the multipli-cation by x:

T (xf(x)) = xT (f)(x) (f ∈ S(R)).

Then there is a function c(x) such that

Tφ(x) = c(x)φ(x) (φ ∈ S(R)).

Proof. It’ll suffice to show that T satisfies the assumptions of Lemma 4. Fix y ∈ R. Letφ ∈ S(R) be such that φ(y) = 0. Then, by Corollary 3 there is ψ ∈ S(R) such that

φ(x) = (x− y)ψ(x).

Hence

T (φ)(x) = T ((x− y)ψ(x)) = (x− y)T (φ)(x),

which is zero if x = y. �

Proposition 6. Suppose T : S(R) → S(R) is a linear map which commutes with themultiplication by x:

T (xf(x)) = xT (f)(x) (f ∈ S(R))

and with the derivative

T (f ′) = T (f)′ (f ∈ S(R)).

Then there is a constant c such that

Tφ(x) = cφ(x) (φ ∈ S(R)).

Proof. We know from Lemma 5. that T coincides with the multiplication by a functionc(x). Since T commutes with the derivative we see that for any f ∈ S(R)

c(x)f ′(x) = (c(x)f(x))′.

Since the multiplication by c(x) prserves the Schwartz space, c(x) is differentiable and

(c(x)f(x))′ = c′(x)f(x) + c(x)f ′(x) .

Hence c′(x) = 0. therefore c(x) is a constant. �

Lemma 7. Let Rf(x) = f(−x). The map T = RF2 : S(R) → S(R) commutes with themultiplication by x and with the derivative.

Proof. Since

R(xf(x)) = −xf(−x) = −xR(f)(x) and R(f ′) = −R(f) ,

it’ll suffice to check that

F2(xf(x)) = −xF2(f)(x) and F2(f ′) = −(F2(f)

)′,

which follows from the relations (3) and (4). �

Lemma 8. Let f(x) = e−πx2. Then Ff = f . (Fourier transform of the normalized

Gaussian is the same Gaussian.)

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Proof. Since

d

dxf(x) = −2πxf(x)

the formulas (3) and (4) show that

d

dx(Ff(x) · f(x)) = 0 .

Hence

Ff(x) = constf(x) .

Evaluating at x = 0 gives ∫Rf(y) dy = const .

By squaring the integral and using polar coordinates we show that const = 1. �

Now we are ready to prove the inversion formula in Theorem 1. We see from that themap RF2 : S(R)→ S(R) is the identity,

RF2f = f (f ∈ S(R)).

We know from Lemma 7 and Proposition 6 that the map RF2 is a constant multiple ofthe identity: cI = RF2. Now Lemma 8 shows that with f(x) = e−πx

2

cf = RF2f = Rf = f.

Thus c = 1. Hence

RF = F−1

and the formula2 follows. This completes the proof of Theorem 1.

Problem 1. Let P (R) denote the space of complex valued polynomials f(x), on the realline. Suppose T : P (R) → P (R) is a linear map that commutes with multiplication by xand with the derivative. Is it true that T is a constant multiple of the identity? If yes,prove it. If no, give a counterexample.

The answer is YES and the proof goes along the same lines we followed for the caseof S(R). Another approach is to define c(x) = T (1) and from the commutation withmultiplication by x deduce that

T (xk) = c(x)xk (k = 0, 1, 2, ... .

Hence

T (f(x)) = c(x)f(x) (f(x) ∈ P (R)) .

Since T commutes with the derivative, we conclude that c′(x) = 0. Therefore c(x) is aconstant.


Problem 2. Show that

F4 = I .

Since

F2 = R−1 = R .

we see that

F4 = R2 = I .

Problem 3. Let γ(x) = e−πx2

and let A = 2πx − ∂x. We already know that Fγ = γ.Check that

F(Akγ) = e−π2ikAkγ (k = 0, 1, 2, 3) .

Thus we found an eigenvector for each of the four possible eigenvalues.

This may be done by an explicit computation using the formulas (3) and (4). For asystematic treatment see [HT92, section 2.1].

Integration by parts shows that for any f ∈ S(R),

F(Af)(y) = −iAF(f)(y) .

By taking f = Akγ we obtain the following recurrence equation

F(Ak+1γ)(y) = −iAF(Akγ) (k = 0, 1, ...) .

Since Fγ = γ, we see that

F(Akγ)(y)) = (−i)kAkγ(y) (k = 0, 1, ...) .

3. Magnetic Resonance Imaging

Suppose a source at s ∈ R is emitting a signal with frequency ks ∈ R and amplitudeA(s). The collective signal received is

B(x) =

∫RA(s)e2πixks ds .

By Fourier inversion,

A(s) = k

∫RB(x)e−2πixks dx .

Hence we can recover A(s) from B(x). In particular, if we the function A(s) is linear (insome large interval contained in [0,∞)) and if we know A(s) then we know s, i.e. thelocation of the source. For the related physics seeyoutube.com/watch?v = pGcZvSG805Y youtube.com/watch?v = djAxjtN 7V E.

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4. The Fourier Transform on the space C∞(U1)

Here U1 = {u ∈ C; |u| = 1} is the group of the unitary matrices of size 1. We shalluse the following identification of groups

R/2πZ 3 θ + 2πZ→ eiθ ∈ U1 .

Here we follow [SS03, section 2.2]. Recall that a function f : Z → C is called rapidlydecreasing if

supn∈Z|(1 + |n|)k|f(n)| <∞ (n ∈ Z) .

Theorem 9. The Fourier transform

f(n) = Ff(n) =

∫ 1

0

e−2πixnf(x) dx (n ∈ Z, f ∈ C∞(U1)) (6)

maps the space C∞(U1) into the space of the rapidly decreasing functions on Z, is invert-ible, and the inverse given by

f(x) =∑n∈Z

e2πixnFf(n) (x ∈ R, f ∈ C∞(U1)) . (7)

f(0) =∑n∈Z

Ff(n) .

Notice that the integral (6) converges as long as the function f is absolutely integrable.Hence we have the Fourier transform Ff for any f ∈ L1(U1). Since∫ 1

0

e−2πixnf ′(x) dx = 2πinFf(y) , (8)

the rapid decrease of Ff is easy to check. The inversion formula is immediate from thefollowing Lemma.

Lemma 10. If f : U1 → C is continuous and Ff = 0, then f = 0.

Proof. Since ∫ 1

0

e−2πixnf(x+ y) dx = e2πiyn

∫ 1

0

e−2πixnf(x) dx

it’ll suffice to show that if Ff = 0, then f(0) = 0.Suppose the claim is false. We may assume that f is real valued and that f(0) > 0.

We shall arrive at a contradiction. Choose 0 < δ ≤ 14

so that

f(x) >f(0)

2(|x| < δ) .

Let ε > 0 be so small that the function

p(x) = ε+ cos(2πx)

satisfies

|p(x)| < 1− ε

2(δ ≤ |x| ≤ 1

2) .


Choose 0 < η < δ so that

|p(x)| ≥ 1 +ε

2(|x| ≤ η) .

Then for k = 0, 1, 2, ...,∣∣∣∣∣∫δ≤|x|≤ 1

2

f(x)p(x)k dx

∣∣∣∣∣ ≤ 2π ‖ f ‖∞(

1− ε

2

)k,∫

η≤|x|<δf(x)p(x)k dx ≥ 0 ,∫

|x|<ηf(x)p(x)k dx ≥ 2η

f(0)

2

(1 +

ε

2

)k.

Therefore

limk→∞

∫|x|≤ 1

2

f(x)p(x)k dx = +∞ .

However p(x)k is a trigonometric polynomial (linear combination of powers of exponen-tials). Hence the assumption Ff = 0 implies∫

|x|≤ 12

f(x)p(x)k dx = 0 .

This is a contradiction we have been looking for. �

Set

en(x) = e2πinx (n ∈ Z, x ∈ R) .

Problem 4. Prove the inclusion

L2(R/Z) ⊆ L1(R/Z) .

This follows from Cauchy’s inequality. Let f ∈ L2(R/Z). Then

∫ 1

0

|f(x)| dx =

∫ 1

0

1 · |f(x)| dx ≤

√∫ 1

0

12 · dx

√∫ 1

0

|f(x)|2 · dx

=

√∫ 1

0

|f(x)|2 · dx <∞ .

For f ∈ L2(R/Z) define the partial Fourier sums

SN(f) =N∑

n=−N

f(n)en (N = 0, 1, 2, ...) .

The crown jewel of the L2 theory for Fourier series is the following theorem.

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Theorem 11. For any f ∈ L2(R/Z)

limN→∞

‖ f − SN(f) ‖2= 0 (9)

and‖ f ‖2

2=∑n∈Z

|f(n)|2 . (10)

Problem 5. Find a prove of Theorem 11 and copy it by hand.

My favorite is [SS03, section 1.2], but there are plenty of other references.

Problem 6. Compute f(n) for f(x) = x, restricted to the interval [−12, 1

2) (and repeated

periodically with period 1).

Integration by parts shows that

f(n) =

{i(−1)n

2πnif n 6= 0 ,

0 if n = 0 .

Problem 7. Compute∞∑n=1

1

n2.

From Problem 6 and Theorem 11,(1

2π

)2 ∞∑n=1

1

n2=

1

2

∑n∈Z\{0}

|f(n)|2 =1

2

∫ 12

− 12

x2 dx =1

24.

Therefore∞∑n=1

1

n2=π2

6.

5. Symmetries of platonic solids

The group S4 acts on R4 by permuting the coordinates and multiplication by the sign ofthe permutation:

R4 3 x = (x1, x2, x3, x4)→ σ x = ((σ x)1, (σ x)2, (σ x)3, (σ x)4) ∈ R4,

(σ x)j = sgn(σ)xσ−1(j) (σ ∈ S4, 1 ≤ j ≤ 4).

Recall the scalar product

(x, y) = x1y1 + x2y2 + x3y3 + x4y4.

It is easy to check that

(σ(x), σ(y)) = x · y (x, y ∈ R4).


Let

V = {x = (x1, x2, x3, x4) ∈ R4;x1 + x2 + x3 + x4 = 0}.

The action of S4 on R4 preserves the subspace V . The vectors

e1 = (−1, 1, 1,−1), e2 = (1,−1, 1,−1), e3 = (1, 1,−1,−1)

form an orthogonal basis of V . Let C be the cube whose faces are centered at point ±ej,1 ≤ j ≤ 3.

Here is a description of how each of the 23 nontrivial elements of S4 acts on the cubeC.

Rotations by 180 degrees about the axis going though midpoints of opposite edges:

(12) : e1 → −e2, e2 → −e1, e3 → −e3,

(13) : e1 → −e3, e2 → −e2, e3 → −e1,

(14) : e1 → −e1, e2 → e3, e3 → e2,

(23) : e1 → −e1, e2 → −e3, e3 → −e2,

(24) : e1 → e3, e2 → −e2, e3 → e1,

(34) : e1 → e2, e2 → e1, e3 → −e3.

Rotations by 120 degrees about the axis going through opposite verticies:

(123) : e1 → e2, e2 → e3, e3 → e1,

(132) : e1 → e3, e2 → e1, e3 → e2,

(134) : e1 → −e2, e2 → −e3, e3 → e1,

(143) : e1 → e3, e2 → −e1, e3 → −e2,

(243) : e1 → −e2, e2 → e3, e3 → −e1,

(234) : e1 → −e3, e2 → −e1, e3 → e2

(124) : e1 → −e3, e2 → e1, e3 → −e2,

(142) : e1 → e2, e2 → −e3, e3 → −e1.

Rotations by 180 degrees about the axis going through the centers of the opposite faces:

(12)(34) : e1 → −e1, e2 → −e2, e3 → e3,

(13)(24) : e1 → −e1, e2 → e2, e3 → −e3,

(14)(23) : e1 → e1, e2 → −e2, e3 → −e3.

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Rotations by 90 degrees about the axis going through the centers of the opposite faces:

(1234) : e1 → e3, e2 → e2, e3 → −e1,

(1324) : e1 → e3, e2 → −e1, e3 → e3,

(1432) : e1 → −e3, e2 → e2, e3 → e1

(1243) : e1 → e1, e2 → −e3, e3 → e2,

(1342) : e1 → e2, e2 → e3, e3 → −e2,

(1423) : e1 → −e2, e2 → e1, e3 → e3.

Let T be a regular tetrahedron whose edges are centered at ±ej, 1 ≤ j ≤ 3. Thenthe subgroup of S4 that preserves T is Alt(4), the subgroup of the even permutations.———–Let Alt(n) ⊆ Sn denote the subgroup consisting of all the even permutations. Then Alt(5)is generated by Alt(4), identified with the subgroup of S5 fixing the number 5, and thecycle s = (12345).

Indeed, Notice that s is an even permutation, i.e. s ∈ Alt(5). We need to show thatfor every g ∈ Alt(5) there is an element h ∈ Alt(4) and an integer k such that

g = hsk. (11)

There is a number x ∈ {1, 2, 3, 4, 5} such that g(x) = 5. Since s−1 is the full cycle, thereis k such that s−k maps 5 to x. Hence the composition h = gs−k maps 5 to 5. Thus hfixes 5 and is an even permutation. In other words, h ∈ Alt(4). Hence (11) follows.

Recall the golden mean µ = 1+√

52

and the conjugate µ = 1−√

52

. In the three-dimensionalspace V = Re1 + Re2 + Re3, let D be the regular dodecahedron whose faces are centeredon the vectors

−µe2 + e3, e1 + µ, e3 − e1 + µe3, −µe1 − e2, −µe2 − e3, µe1 − e2 (12)

and on the antipodal vectors. The action of Alt(4) preserves the dodecahedron (i.e. itpreserves the set of vectors (12) multiplied by ±1).

Indeed, leta = (12)(34), , b = (14)(23), c = (13)(24),

x = (123), y = (142), z = (243), w = (134).

The group Alt(4) consists of

1, a, b, c, x, x2, y, y2, z, z2, w, w2.

Since,xax−1 = b, x2ax−2 = c, axa = y, bxb = z, cxc = w,

we see that Alt(4) is generated by a and x. Hence, it’ll suffice to check that a and xpreserve the dodecahedron.


LetF1 = −µe2 + e3, F2 = e1 + µe3, F3 = −e1 + µe3,

F4 = −µe1 − e2, F5 = −µe2 − e3, F6 = µe1 − e2

These are the centers of the 6 faces of the dodecahedron. The centers of the antipodalfaces are

−F1, −F2, −F3, −F4, −F5, −F6.

We see from the explicit action of the group that

a : e1 → −e1, e2 → −e2, e3 → e3

and

x : e1 → e2, e2 → e3, e3 → e1.

Hence,

a : F1 → −F5, F2 → F3, F3 → F2, F4 → −F4, F5 → −F1, F6 → −F6,

and

x : F1 → −F3, F2 → −F4, F3 → F6, F4 → F5, F5 → −F2, F6 → −F1.

By computing the angles between the vectors we see that the face F1 shares one edgewith each of the faces F2, F3, F4, F5, F6. Furthermore, F2 shares one edge with F3, F3

shares one edge with F4, F4 shares one edge with F5, F5 shares one edge with F6 and F6

shares one edge with F2. This determines the model of the dodecahedron and we checkthat both a and x preserve it.

We extend the action of Alt(4) to an action of Alt(5) on V by letting s by

s : −µe2 + e3 → −µe2 + e3

s : e1 + µe3 → −e1 + µe3 → −µe1 − e2 → −µe2 − e3 → µe1 − e2 → e1 + µe3.

Then the action of s is the rotation by 2π/5 about the axis passing through midpoints−µe2 + e3 and µe2 − e3 of the two opposite faces. (This way Alt(5) is identified with thegroup of the rotations of a regular dodecahedron.)

Indeed, in terms of the notation used above, the action of s looks as follows:

s : F1 → F1,

s : F2 → F3 → F4 → F5 → F6 → F2.

We may construct a model of the dodecahedron using, for example a suitable link on theweb. Furthermore, since we may compute all the angles between the vectors we knowwhere to place the faces F1, F2, F3, F4, F5, F6, F2. After all that it is clear that s has thedesired property.

14 TOMASZ PRZEBINDA

The matrix of s with respect to the basis {e1, e2, e3} is

1

2

−µ −1 −µ1 µ µµ µ 1

.

Indeed, We see from the definition of s that

s : e1 =1

2(F2 − F3)→ 1

2(F3 − F4) =

1

2((−1 + µ)e1 + e2 + µe3)

s : e2 = −1

2(F4 + F6)→ −1

2(F5 + F2) =

1

2(−e1 + µe2 − (µ− 1)e3)

s : e3 =1

2(F1 − F5)→ 1

2(F1 − F6) =

1

2(−µe1 + (1− µ)e2 + e3).

Since, 1− µ = µ the formula for the matrix follows.

6. Basic representation theory of finite groups

A good reference for this section is [FH91]. A representation of a finite group G on afinite dimensional complex vector space V is a group homomorphism

ρ : G→ GL(V). (13)

We shall denote it by (ρ,V) or just ρ. Also, when convenient we shall write gv instead ofρ(g)v. A morphism, or a G-morphism, or a G-intertwinig map, between two representa-tions (ρ,V) and (ρ′,V′) is a linear map T : V→ V′ such that

Tρ(g)v = ρ′(g)Tv (g ∈ G, v ∈ V).

Two representations (ρ,V) and (ρ′,V′) are called isomorphic if there is an invertible mor-phism between them. In that case we shall write (ρ,V) ' (ρ′,V′). If the representationsare not isomorphic we shall write (ρ,V) 6' (ρ′,V′). The relation of an isomorphism ofrepresentations is an equivalence relation.

Let Vc denote the vector space dual to V. The contragredient representation (ρc,Vc) isdefined by

ρc(g)vc(v) = vc(ρ(g−1)v) (v ∈ V, vc ∈ Vc, g ∈ G).

Given two representations (ρ,V) and (ρ′,V′) define their direct sum (ρ⊕ ρ′,V ⊕ V′) by

(ρ⊕ ρ′)(g)(v, v′) = (ρ(g)(v), ρ′(g)(v′)) (g ∈ G, v ∈ V, v′ ∈ V′)

and the tensor product (ρ⊗ ρ′,V ⊗ V′) by

(ρ⊗ ρ′)(g)[v ⊗ v′] = [ρ(g)(v)]⊗ [ρ′(g)(v′)] (g ∈ G, v ∈ V, v′ ∈ V′).


Given a representation (ρ,V), one says that a subspace U ⊆ V is invariant if ρ(g)u ∈ U forall u ∈ U. A representation (ρ,V) is called irreducible if the only G-invariant subspacesof V are 0 and V.

Theorem 12. (Schur’s lemma) If (ρ,V) and (ρ′,V′) are two irreducible representationsof G and T : V → V′ is a morphism, then either T is an isomorphism or T = 0. If(ρ,V) = (ρ′,V′) then T is the multiplication by a complex number.

Proof. Notice thet Ker(T ) ⊆ V is a G-invariant subspace. Hence, either Ker(T ) = 0 orKer(T ) = V. Similarly, the image of T , Im(T ) ⊆ V′ is a G-invariant subspace. Hence,either Im(T ) = 0 or Im(T ) = V. Thus the first statement follows.

The second statement follows from the fact tht C is algebraically closed. The point isthat T has an eigenvalue λ ∈ C and the corresponding (non-zero) eigen-space in V. Byirreducibility this eigenspace is equal to V. �

Let HomG(V,V′) denote the space of all the morphisms fro V to V′. Theorem 12 impliesthe following Corollary

Corollary 13. For two irreducible representations (ρ,V) and (ρ′,V′) of G,

dim HomG(V,V′) =

{1 if (ρ,V) ' (ρ′,V′)0 if (ρ,V) 6' (ρ′,V′)

Lemma 14. For any finite dimensional representation (ρ,V) there is a G-invariant posi-tive definite hermitian form ( , ) on V.

Proof. Let 〈 , 〉 be any positive definite hermitian form on V. Define

(u, v) =∑g∈G

〈ρ(g)u, ρ(g)v〉 (u, v ∈ V).

It is easy to check that this form has the required properties. �

If V is equipped with an invariant positive definite hermitian form then (ρ,V) is calleda unitary representation of G.

Theorem 15. Any finite dimensional representation of a finite group decomposes into thedirect sum of irreducible representations.

Proof. This follows from Lemma 14 and the fact that the orthogonal complement of aG-invariant subspace is G-invariant. �

Two unitary representations (ρ,V), (ρ′,V′) are called unitarily equivalent iff there is amorphism T : V→ V′ which is also an isometry.

Proposition 16. If two unitary representations (ρ,V), (ρ′,V′) are equivalent then they areunitarily equivalent.

Problem 8. Proof Proposition 40.

16 TOMASZ PRZEBINDA

Proof. Let ( , ) be the invariant scallar product on V and let ( , )′ be the invariant scallarproduct on V′. Pick an isomorphism T : V→ V′. Define T ∗ : V′ → V by

(Tu, v)′ = (u, T ∗v) (u ∈ V, v ∈ V).

Then T ∗V′ → V is also a morphism. Hence, T ∗T : V → V commutes with the action ofG. Hence, there is λ ∈ C such that T ∗T = λI. Thus, for any u, v ∈ V,

(Tu, Tv)′ = (u, T ∗Tv) = λ(u, v).

In particular, by taking u = v 6= 0 we see that λ > 0. Hence,

1√λT : V→ V′

is an isometry and a morphism. �

It is easy to see from the above theorem that any irreducible representation of a finiteabelian group is one-dimensional. Indeed, let G be such a group and let (ρ,V) be anirreducible unitary representation of G. For a fixed g ∈ G, the map ρ(g) : V → Vintertwines the action of G. Theorem 12 implies that there is ch(g) ∈ C such thatρ(g) = χ(g)IV. Since ρ(g) is invertible, we see that χ(g) 6= 0. Moreover the map G 3 g →χ(g) ∈ C× is a group homomorphism, i.e. a character. Therefore every subspace U ⊆ Vis G-invariant. Since (ρ,V) is irreducible we see that dimV = 1.

Any finite abelian group is isomorphic to the direct product of cyclic groups and thecharacters of cyclic groups are easy to describe. Hence to make things interesting weneed a non-abelian group. The simplest one is S3 the group of all the permutations of 3elements.

As our first example we shall describe all the irreducible representations of S3. Thereare two obvious one-dimensional representations: the trivial representation (triv,C) andthe sign representation (sgn,C):

triv(g)v = v, sgn(g)v = the sign of the permutation g times v (g ∈ S3, v ∈ C).

There is also a two-dimensional irreducible representation (ρ,V) constructed as follows.Define a representation (π,C3) by

π(g)(z1, z2, z3) = (zg−1(1), zg−1(2), zg−1(3)) (g ∈ S3, (z1, z2, z3) ∈ C3).

This is a unitary representation with respect to the scalar product

((z1, z2, z3), (z′1, z′2, z′3)) = z1z′1 + z2z′2 + z3z′3.

The orthogonal complement U ⊆ C3 of C(1, 1, 1) is S3-invariant and two dimensional. Letρ denote the restriction of π to U. Then (ρ,U) is a representation of S3.

Let c = (123) ∈ S3, t = (12) ∈ S3, z = e2πi/3 ∈ C, u = (z, 1, z2) ∈ C3 and letv = (1, z, z2) ∈ C3. Then the vectors u and v form a basis of the vector space U and

ρ(c)u = zu, ρ(c)v = z2v, ρ(t)u = v, ρ(t)v = u. (14)


In particular we see thet U does not have any S3-invariant one-dimensional subspace.Thus the representation (ρ,U) is irreducible. We claim that up to equivalence there areno other irreducible representations of S3.

Indeed, consider a finite dimensional irreducible representation (ρ,V) of S3. SupposedimV = 1. Then ρ : S3 → GL(V) = C× is a group homomorphism. Hence, the kernelKer(ρ) ⊆ S3 is a normal subgroup. Since the group C× is abelian, Ker(ρ) 6= {1}. IfKer(ρ) = S3, then ρ is the trivial representation. If Ker(ρ) = {1, c, c2}, the subgroupof cycles of length 3, then ρ is the sign representation. Since there are no other normalsubgroups in S3, we see that there are no other one-dimensional representations of S3.

Suppose dimV ≥ 2. Then the subgroup {1, c, c2} ⊆ S3, of index 2, cannot act triviallyon V, because the quotient S3/{1, c, c2} is abelian. Hence there is a vector u ∈ V suchthat ρ(c)u = λu, where λ = e2πi/3 or e−2πi/3. Notice that ct = tc2. Hence

ρ(c)ρ(t)u = tc2u = ρ(t)(λ2u) = λ2ρ(t)u.

Therefore we may assume that λ = e2πi/3 = z in our previous notation (14). Let v = ρ(t)u.Then, from the above computation, tv = λ2v. We see from the above and from (14) thatthe vectors u and v span a two dimensional subspace isomorphic to the representation onthe space U constructed before. Since V is assumed irreducible, that subspace coincideswith V. Thus (ρ,V) is isomorphic to (ρ,U).

Problem 9. Read the newly added section 5 and correct errors if you find any.

There was µ in the center of the matrix of s. In the first line of “Rotations by 180degrees about the axis going though midpoints of opposite edges”, in the first line weshould have e3 → −e3.

Now we come back to general representation theory of a finite group G. By definition thecharacter χV = χρ of a representation (ρ,V) is the following complex valued function onthe group:

χV(g) = tr(ρ(g)) (g ∈ G).

This function is invariant under conjugation

χ(hgh−1) = χ(g) (h, g ∈ G).

Also, we haveχV⊕V′ = χV + χV′ , χV⊗V′ = χVχV′ and χρc = χρ. (15)

(The last formula follows from the fact that the eigenvalues of ρ(g) are nth roots of unity,where n = |G|, and therefore the inverse of such an eigenvalue is the same as its complexconjugate.)

Theorem 17. Let X be a set and let V be the space of all the complex valued functionsdefined on X. Suppose G acts on X by permutations. Define a representation (ρ,V) by

ρ(g)v(x) = v(g−1x) (g ∈ G, v ∈ V, x ∈ X).

Thenχρ(g) = the number of points in X which are fixed by g.

18 TOMASZ PRZEBINDA

Proof. Let δy denote the Dirac delta at y ∈ X. The set of all these Dirac deltas is a basisof the vector space V. Notice that

ρ(g)δy = δgy (g ∈ G, y ∈ X).

Let us order the set X as X = {y1, y2, ..., yn}. The matrix [A(g)j,k] of ρ(g) with respectto the order basis {δy1 , δy2 , ..., δyn} are computed from the formula

ρ(g)δyk =n∑j=1

A(g)j,kδyj .

Explicitly

δgyk =n∑j=1

A(g)j,kδyj .

Thus A(g)j,k is either 0 or 1, and later happens if and only if δgyk = δyj . In particularA(g)j,j = 1 if and only if δgyj = δyj . Hence the formula for the character follows. �

As an application we compute the characters of the irreducible representations of S3 inthe following table. The characters for the trivial and the sign representation are obvious.We see from the definition of the two-dimensional irreducible representation and from (24)that its character is equal to the difference between the character χC3 of the permutationrepresentation of S3 on C3 and the trivial representation. Theorem 17 that

χC3(1) = 3, χC3((12)) = 1, χC3((123)) = 0.

Hence the last line of the table follows.

conjugacy class in S3 [1] [(12)] [(123)]

number of elements in the conjugacy class 1 3 2

value of the character on the conjugacy class for

trivial representation 1 1 1

sign representation 1 −1 1the two-dimensional representation 2 0 −1

We introduce the following scalar (hermitian) product on the space of all the complexvalued functions on G

(φ, ψ) =1

|G|∑g∈G

φ(g)ψ(g). (16)

In other words we view the counting measure divided by the cardinality of the group asa fixed Haar measure on G and we consider the corresponding space L2(G). Denote byL2(G)G ⊆ L2(G) the subspace of the functions invariant by the conjugation by all theelements of G. Our characters live in L2(G)G.


Lemma 18. Suppose (ρ,V) is a non-trivial irreducible representation of G. Then∑g∈G

ρ(g)v = 0 (v ∈ V).

Proof. Notice that

v =∑g∈G

ρ(g)v

is a G-invariant vector in V. Since ρ is irreducible, either v = 0 or V = Cv. Since ρ isnon-trivial, the second option is impossible. �

Corollary 19. Suppose (ρ,V) is a non-trivial irreducible representation of G. Then∑g∈G

χρ(g) = 0.

Corollary 20. Suppose (ρ,V) is a representation of G. Then

1

|G|∑g∈G

χρ(g) = dimVG,

where VG ⊆ V is the space of the G-invariant vectors.

Lemma 21. The characters of irreducible representations form an orthonormal set inL2(G)G.

Proof. Consider two such representations (ρ,V) and (ρ′,V′). Let Hom(V,V′) denote thespace of all the linear maps from V to V′. The group G acts on this vector space by

gT (v) = ρ′(g)Tρ(g−1)v (g ∈ G, T ∈ Hom(V,V′), v ∈ V).

This way Hom(V,V′) becomes a representation of G. It is easy to check that as such it isisomorphic to (ρc ⊗ ρ′,Vc ⊗ V′). Hence, by (24),

χHom(V,V′)(g) = χρ(g)χρ′(g) (g ∈ G).

Therefore,

(χ′ρ, χρ) =1

|G|∑g∈G

χHom(V,V′)(g)

= dimension of the space of the G-invariants in Hom(V,V′).

Thus the formula follows from Corollary 39. �

Let (ρ,V) be a representation of G and let

V = V1 ⊕ V2 ⊕ ...⊕ Vn

be the decomposition into irreducibles. The number of the Vj which are isomorphic toV1 is called the multiplicity of V1 in V, denoted m1. We may collect the isomorphic

20 TOMASZ PRZEBINDA

representations in the above formula and (after changing the indecies appropriately) getthe following decomposition

V = V⊕m11 ⊕ V⊕m2

2 ⊕ ...⊕ V⊕mkk ,

where the Vj are irreducible and mutually non-isomorphic and

V⊕mjj = Vj ⊕ Vj ⊕ ...⊕ Vj (mj summands).

This is the primary decomposition of V, which is also denoted by

V = m1V1 ⊕m2V2 ⊕ ...⊕mkVk = m1 · V1 ⊕m2 · V2 ⊕ ...⊕mk · Vk,for brevity.

Proposition 22. LetV = m1V1 ⊕m2V2 ⊕ ...⊕mkVk,

be the primary decomposition of V. Then

χV = m1χV1 +m2χV2 + ..mkχVk

andmj = (χV, χVj).

In particular V is irreducible if and only if (χV, χV) = 1.

Proof. This follows from (24) and from Lemma 49. �

Let (L,L2(G)) be the left regular representation of G:

L(g)φ(h) = φ(g−1h) (g, h ∈ G, φ ∈ L2(G)).

Theorem 23. The character of the left regular representation is given by

χL(g) =

{|G| if g = 1,0 if g 6= 1.

(17)

Let (ρ1,V1), (ρ2,V2), ..., (ρk,Vk) be the set of all (up to equivalence) irreducible represen-tations of G. (Lemma 49 shows that this set is finite.) Then

χL = dimV1 χV1 + dimV2 χV2 + ...dimVk χVk ,

or equivalently,

L2(G) = dimV1 · V1 ⊕ dimV2 · V2 ⊕ ...⊕ dimVk · Vk.In particular every irreducible representation of G occurs in L2(G). Also,

|G| = (dimV1)2 + (dimV2)2 + ...+ (dimVk)2.

Proof. The formula for χL follows from Theorem 17. Suppose a representation (ρj,Vj)occurs in the left regular representation. Then the multiplicity of Vj in L2(G) is equal to

(χρj , χL) =1

|G|∑g∈G

χρj(g)χL(g) = χρj(1) = dimVj, ,


where the second equality follows from (17). Since we already have Lemma 49, it remainsto check that every irreducible representation (ρj,Vj) occurs in (L,L2(G)). For two anon-zero vector u, v ∈ Vj define the corresponding matrix coefficient of the irreduciblerepresentation

Mu,v(g) = (u, ρj(g)v) (g ∈ G).

It is easy to check that the map

Vj 3 u→Mu,v ∈ L2(G)

is a morphism. Since this map is non-zero, we see that (ρj,Vj) occurs in (L,L2(G)). �

Problem 10. Describe the irreducible representations of the group S4 and their characters.

There are the trivial and the sign representations, as in the case of S3. The group S4 actson C4 and the orthogonal complement of the trivial representation (U = C(1, 1, 1, 1)⊥ ⊆C3) provides a 3 dimensional representation. Its character may be computed from Theo-rem 17. Then one checks that (χU, χU) = 1. Hence Proposition 51 shows that U is irre-ducible. Let U′ stand for the representation obtained from U via tensoring with the signrepresentation. Then U′ is also irreducible (for the same reason). The squares of the di-mensions of the representations we constructed so far add up to 12+12+32+32 = 20. Hencewe know from Theorem 23 that we’ll be done if we find an irreducible two-dimensionalrepresentation, call it W. In fact, since the characters of the irreducible representationsform an orthonormal basis of L2(G)G it suffices to find a function χW ∈ L2(G)G which isorthogonal to all the characters we found so far, has norm 1 (χW, χW) = 1) and χW(1) = 2.This can be done, of course. We summarize the results in the following table.

conjugacy class in S4 [1] [(12)] [(123)] [(1234)] [(12)(34)]number of elements in the conjugacy class 1 6 8 6 3


trivial representation 1 1 1 1 1sign representation 1 −1 1 −1 1

the three-dimensional representation U 3 1 0 −1 −1U tensored with sign 3 −1 0 1 −1

the two-dimensional representation W 2 0 −1 0 2

(Notice that {1, (12)(34), (13)(24), (14)(23)} ⊆ S4 is a normal subgroup. Then one checksthat W is the pullback of the irreducible two-dimensional representation of S3 to S4 viathe quotient map

S4 → S4/{1, (12)(34), (13)(24), (14)(23)} = S3.)

The alternating group A4 ⊆ S4 contains {1, (12)(34), (13)(24), (14)(23)} as a normal sub-group and the quotient is a cyclic group of order 3. Hence, A4 has 3 one-dimensionalrepresentations. Here is the charter table for A4, where z = e2πi/3.

22 TOMASZ PRZEBINDA

conjugacy class in A4 [1] [(123)] [(132)] [(12)(34)]number of elements in the conjugacy class 1 4 4 3


trivial representation 1 1 1 1second one dimensional representation 1 z z2 1third one dimensional representation 1 z2 z 1the three-dimensional representation 3 0 0 −1

Lemma 24. Given an irreducible unitary representation (ρ,V) of G, consider End(V) asa Hilbert space with the inner product

(S, T ) = tr(ST ∗) (S, T ∈ End(V)) .

Let

π(g1, g2)T = ρ(g2)Tρ(g−11 ) (g1, g2 ∈ G, T ∈ End(V)) .

Then (π,End(V)) is an irreducible unitary representation of G×G.

Proof. The non-trivial part is to check the irreducibility. Notice that (π,End(V)) is iso-morphic to outer tensor product (ρc ⊗ ρ,Vc ⊗ V). Hence,

1

|G×G|∑

(g1,g2)∈G×G

|χπ(g1, g2)|2 =1

|G×G|∑

(g1,g2)∈G×G

|χρ(g1)χρ′(g2)|2

=1

|G|∑g1∈G

|χρ(g1)|2 1

|G|∑g2∈G

|χρ′(g2)|2 = 1.

Hence, by Proposition 51, (π,End(V)) is irreducible. �

Given a representation (ρ,V) define a map

Mρ : End(V)→ L2(G), Mρ(T )(g) = tr(ρ(g)T ) = tr(Tρ(g)).

In particular, if I ∈ End(V) is the identity, then

Mρ(I) = χρ.

Also, it is easy to see that the subspace Mρ(End(V)) ⊆ L2(G) depends only on theequivalence class of ρ.

Recall the right regular representation of G,

R(g)f(h) = f(hg) (g, h ∈ G, f ∈ L2(G)). (18)

and the outer tensor product L⊗R,

L⊗R(g1, g2)f(g) = f(g−11 gg2) (g1, g2, g ∈ G, f ∈ L2(G)) . (19)

It is easy to see that

L(g1)Mρ(T ) = Mρ(Tρ(g−11 )) , R(g2)Mρ(T ) = Mρ(ρ(g2)T ) (g1, g2 ∈ G) . (20)


In particular,

Mρ(T )(g−1hg) = Mρ(ρ(g)Tρ(g)−1)(h) (g, h ∈ G) . (21)

Lemma 25. Given an irreducible unitary representation (ρ,V) of G, The map

Mρ : End(V)→ L2(G)

intertwines the representation π with the outer tensor product L⊗R, is injective, and

1

dimVtr(ST ∗) = (Mρ(S),Mρ(T )) (S, T ∈ End(V)) .

Proof. The intertwining property follows from (20). Therefore the irreducibility of πimplies the injectivity of Mρ.

Recall the adjoint map between Hilbert spaces,

M∗ρ : L2(G)→ End(V).

Notice also that

M∗ρMρ : End(V)→ End(V)

is G×G-intertwining. Hence there is λ ∈ C such that M∗ρMρ = λI. Furtheremore,

λ · dimV = (I,M∗ρMρ(I)) = (Mρ(I),Mρ(I)) = (χρ, χρ) = 1

and we are done. �

Theorem 26. Suppose (ρ,V) and (ρ′,V′) are two irreducible unitary representations of G.Then

a) if (ρ,V) is not equivalent to (ρ′,V′), then Mρ(End(V)) ⊥Mρ′(End(V′)),b) for any S, T ∈ End(V), (Mρ(S),Mρ(T )) = 1

dimVtr(ST ∗).

Proof. We have the adjoint map

M∗ρ′ : L2(G)→ End(V′).

Notice also that

M∗ρ′Mρ : End(V)→ End(V′)

is a morphism (i.e. G×G intertwining map). If the two representations are not isomorphicthen, by Theorem 12, M∗

ρ′Mρ = 0. Thus for S ∈ End(V) and T ∈ End(V′),

(Mρ(S),Mρ′(T )) = (S,M∗ρ′Mρ(T )) = 0

and a) follows. Part b) was verified in Lemma 25. �

Theorem 27. There are the following direct sum orthogonal decompositions of Hilbertspaces

a) L2(G) =∑

ρ∈G Mρ(End(V)),

b) L2(G)G =∑

ρ∈G Cχρ.Thus the characters χρ form an orthonormal basis of L2(G)G.

24 TOMASZ PRZEBINDA

Proof. Part a) follows from Theorem 52 by counting dimensions. Part b) follows frompart a), because, by Theorem 12, End(V)G = CI. Indeed,

L2(G)G =∑ρ∈G

(Mρ(End(V)))G =∑ρ∈G

Mρ(End(V)G) =∑ρ∈G

Cχρ.

where only the middle equation

(Mρ(End(V)))G = Mρ(End(V)G)

requires an explanation. We need to show that if

tr(Tρ(ghg−1)) = tr(Tρ(h))

for all h ∈ G, then ρ(g−1)Tρ(g) = T . Since tr(Tρ(ghg−1) = tr(ρ(g−1)Tρ(g)ρ(h)), we’llbe done as soon as we show that if tr(Sρ(h)) = 0 for all h ∈ G, then S = 0. But, sincethe map Mρ is injective (because π is irreducible), this is indeed the case. �

Corollary 28.

δ =∑ρ∈G

χρ(1)

|G|· χρ.

Proof. We see from part b) of Theorem 53 that there are numbers mρ ∈ C such that

δ =∑ρ∈G

mρ · χρ.

But Lemma 49 shows that mρ = (δ, χρ) = χρ(1)

|G| . �

For a function f ∈ L1(G) define the Fourier transform

f(ρ) =1

|G|∑g∈G

f(g)ρc(g) =1

|G|∑g∈G

f(g)ρ(g−1) (ρ ∈ G). (22)

Thus for a representation (ρ,V), f(ρ) ∈ End(V).

Theorem 29. (Fourier inversion for G) For any f ∈ L1(G),

f(g) =∑ρ∈G

χρ(1) · tr(f(ρ)ρ(g)) (g ∈ G) .

In particular,

f(1) =∑ρ∈G

χρ(1) · 1

|G|∑g∈G

f(g)χρc(g) .


Proof. This is straightforward:

f(g) =∑h∈G

f(h)δ(h−1g) =∑h∈G

f(h)

∑ρ∈G

χρ(1)

|G|· χρ

(h−1g)

=∑h∈G

f(h)∑ρ∈G

χρ(1)

|G|· χρ(h−1g) =

∑h∈G

f(h)∑ρ∈G

χρ(1)

|G|· trρ(h−1g)

=∑ρ∈G

χρ(1)

|G|· tr

(∑h∈G

f(h)ρ(h−1g)

)

=∑ρ∈G

χρ(1) · tr

(1

|G|∑h∈G

f(h)ρ(h−1)ρ(g)

).

�

For two functions f1, f2 ∈ L1(G) define the convolution

f1 ∗ f2(g) = L(f1)f2(g) =1

|G|∑h∈G

f1(h)L(h)f2(g) =1

|G|∑h∈G

f1(h)f2(h−1g) (g ∈ G).

Also, for f ∈ L1(G) define

f ∗(g) = f(g−1) (g ∈ G).

Lemma 30. The following formulas hold:

a) for two functions f1, f2 ∈ L1(G), (f1 ∗ f2) = f1f2,

b) for a function f ∈ L1(G), (f ∗) = f ∗.

Proof. This is straightforward:

(f1 ∗ f2)(ρ) =1

|G|∑g∈G

f1 ∗ f2(g)ρc(g)

=1

|G|2∑h∈G

∑g∈G

f1(h)f2(h−1g)ρc(g)

=1

|G|2∑h∈G

∑g∈G

f1(h)ρc(h)f2(h−1g)ρc(h−1g)

=1

|G|2∑h∈G

∑g∈G

f1(h)ρc(h)f2(g)ρc(g) = f1(ρ) · f2(ρ)

26 TOMASZ PRZEBINDA

and

(f ∗)(ρ) =1

|G|∑g∈G

f ∗(g)ρc(g) =1

|G|∑g∈G

f(g−1)ρc(g)

=1

|G|∑g∈G

f(g)ρc(g)∗

=

(1

|G|∑g∈G

f(g)ρc(g)

)∗= f(ρ)∗.

�

Theorem 31. For any f ∈ L2(G),

‖ f ‖2=∑ρ∈G

χρ(1)· ‖ f(ρ) ‖2 .

Proof. We see from Lemma 30 that the left hand side is equal to

f ∗ f ∗(1) =∑ρ∈G

χρ(1) · tr(f(ρ)f(ρ)∗),

which coincides with the right hand side. �

Problem 11. Read the proof of Theorem 56

7. Some Functional Analysis on a Hilbert space

Here we follow [Lan85, sections 1.2 and 1.3]. Let V be a Hilbert space. A continuouslinear map A : V → V is called compact if it maps any bounded sequence vn ∈ V to asequence Avn that has a convergent subsequence.

Theorem 32. Let A be a compact hermitian operator on the Hilbert space V. Thenthe family of eigenspaces Vλ, where λ ranges over all eigenvalues (including 0), is anorthogonal decomposition of E:

V =⊕λ

Vλ .

For λ 6= 0 the eigenspace Vλ is finite dimensional.

Let S be a set of operators (continuous linear maps) on V. We say that V is S-irreducibleif V has no closed S-invariant subspace other than {0} and V itself. We say that V iscompletely reducible for S if V is the orthogonal direct sum of S-irreducible subspaces.Two subspaces V1,V2 ⊆ E are called S-isomorphic if there is an isometry from V1 ontoV2 which intertwines the action of S. The number of elements in such an isomorphismclass is called the multiplicity of that isomorphism class in V.

A subalgebra A of operators on V is said to be ∗-closed if whenever A ∈ A, thenA∗ ∈ A. As explained in [Lan85, section 1.2], the following Theorem is a consequence ofTheorem 32.


Theorem 33. Let A be a ∗-closed subalgebra of compact operators on a Hilbert space V.Then V is completely reducible for A, and each irreducible subspace occurs with finitemultiplicity.

The following theorem (and the corollary below) is known as Schur’s Lemma. For aproof see [Lan85, Appendix 1, Theorem 4]

Theorem 34. Let S be a set of operators acting irreducibely on the Hilbert space V. LetA be a hermitian operator such that AB = BA for all B ∈ S. Then A = cI for some realnumber c.

Corollary 35. Let S be a set of operators acting irreducibely on the Hilbert space V. LetA be an operator such that AB = BA and A∗B = BA∗ for all B ∈ S. Then A = cI forsome c ∈ C. number c.

Proof. We write A = B + iC, where

B =1

2(A+ A∗) , C =

1

2i(A− A∗) .

Then B and C commute with S. Hence there are real numbers b and c such that B = bIand C = cI. Hence A = (b+ ic)I. �

We shall also use a few facts concerning integral kernel operators. The proofs may befound in [Lan85, section 1.3]

Theorem 36. Let (X,M, dx) and (Y,N, dy) be measured spaces, and assume that L2(X),L2(Y ) have countable orthogonal bases. Let q ∈ L2(X × Y ). Then the the formula

Qf(x) =

∫Y

q(x, y)f(y)dy

defines a bounded, compact operator from L2(Y ) into L2(X) with ‖ Q ‖≤‖ q ‖2.

Let A be a bounded operator on a Hilbert space V. We say that A is a Hilbert-Schmidtoperator if there is an orthonormal basis v1, v2, ... of V such that

∞∑n=1

‖ Avn ‖2<∞ .

One checks that tjhis quantity does not depend on the orthonormal basis and denotes thesquare root of it by ‖ A ‖H.S. This is the Hilbert-Schmidt norm of A.

A bounded operator T on a Hilbert space V is of trace class if it is the product of twoHilbert-Schmidt operators, T = AB. Then for any orthonormal basis vn of V∑

n

|(Tvn, vn)| =∑n

|(Bvn, A∗vn)| 5‖ A ‖H.S‖ B∗ ‖H.S=‖ A ‖H.S‖ B ‖H.S<∞ .

Therefore the following series

trT =∑n

(Tvn, vn)

is converges and defines the trace tr of T . For all tht see [Lan85, section VII.1].

28 TOMASZ PRZEBINDA

Also, we have the following theorem, see [Kna86, Lemma 10.15] or [Lan85, Theorem 1,page 128] for a special X,

Theorem 37. Let X be a compact smooth manifold and let dx be a measure on X thatis a smooth function times the Lebesgue measure in each coordinate neighborhood. Letq ∈ C∞(X ×X). Then the the formula

Qf(x) =

∫X

q(x, y)f(y)dy

defines a bounded operator of trace class on L2(X) and

trQ =

∫X

q(x, x) dx .

Problem 12. Get familiar with the notions and theorems of section 7.

8. Representations of locally compact groups

Let G be a locally compact group and let V be a topological vector space. We shallalways assume that V 6= {0}. Let GL(V) denote the group of the invertible continuousendomorphisms of V. A representation of G on V is a pair (π,V), where π : G→ GL(V)is a group homomorpism such that the map

G× V 3 (g, v)→ π(g)v ∈ V

is continuous. A subspace W ⊆ V is called invariant if ρ(G)W ⊆ W. The representation(ρ,V) is called irreducible if V does not contain any closed invariant subspaces other than{0} an V.

If (π1,V1), (π2,V2) are representations of G, then a continuous linear map T : V1 → V2

such thatTπ1(g) = π2(g)T (g ∈ G)

is called an intertwining operator, or a G-homomorphism. The vector space of all the in-tertwining maps will be denoted HomG(V1,V2) or more precisely, HomG((π1,V1), (π2,V2)).The representations (π1,V1), (π2,V2) are equivalent if there exists a bijective T ∈ HomG(V1,V2).

A basic example of a representation of G is the right regular representation (R,C(G).Here C(G) is the space of the continuous, complex valued, functions on G (with thetopology of uniform convergence on compact sets) and

R(g)f(x) = f(xg) (g, x ∈ G, f ∈ C(G)) .

Similarly, we have the left regular representation (L,C(G)),

L(g)f(x) = f(g−1x) (g, x ∈ G, f ∈ C(G)) .

Given an irreducible representation (π,V) and an element λ ∈ V′ (a continuous linearfunctional on V) the formula

Tλv(x) = λ(π(x)v) (x ∈ G)

defines a map Tλ ∈ HomG((π,V), (R,C(G))). If V has enough continuous linear function-als, in the sense that for every 0 6= v ∈ V there is λ ∈ V′ such that λ(v) 6= 0, then Tλ is


injective, so (π,V) may be viewed as a subrepresentation of (R,C(G)). Recall, [Rud91,section 1.47], that for V = Lp([0, 1]) with 0 < p < 1 the dual space V′ = {0}. Thus forthis space the above argument wouldn’t work.

The function

G 3 x→ λ(π(x)v) ∈ Cis called a matrix coefficient of the representation (π,V).

If V is a Hilbert space, then a representation (π,V) is called unitary if every operatorπ(g), g ∈ G, is unitary. Two unitary representations (π1,V1), (π2,V2) are called unitariliequivalen if there is a bjiective and isometric G-homomorphism T : V1 → V2.

Here is the classical version of Schur’s Lemma, which is an immediate consequence ofCorollary 35.

Theorem 38. Let (π,V) be an irreducible unitary representation of G. Then

HomG(V,V) = CI .

Corollary 39. For two irreducible unitary representations (ρ,V) and (ρ′,V′) of G,

dim HomG(V,V′) =

{1 if (ρ,V) ' (ρ′,V′)0 if (ρ,V) 6' (ρ′,V′)

(Here ' stands for unitary equivalence.)

Proposition 40. If two unitary representations (π,V), (π′,V′) are equivalent then theyare unitarily equivalent.

Proof. Let ( , ) be the invariant scallar product on V and let ( , )′ be the invariant scallarproduct on V′. Pick an isomorphism T : V→ V′. Define T ∗ : V′ → V by

(Tu, v)′ = (u, T ∗v) (u ∈ V, v ∈ V).

Then T ∗V′ → V is also a morphism. Hence, T ∗T : V → V commutes with the action ofG. Hence, there is λ ∈ C such that T ∗T = λI. Thus, for any u, v ∈ V,

(Tu, Tv)′ = (u, T ∗Tv) = λ(u, v).

In particular, by taking u = v 6= 0 we see that λ > 0. Hence,

1√λT : V→ V′

is an isometry and a morphism. �

Problem 13. Define the Heisenberg group

H = R× R× R

with the multiplication

(x, ξ, t)(x′, ξ′, t′) = (x+ x′, ξ + ξ′, t+ t′ +1

2(x · ξ′ − x′ · ξ))

30 TOMASZ PRZEBINDA

The Schwartz space S(R) is equipped with a topology, [Hor83, section 7.1]. One may checkthat the formula

(ρ(x, ξ, t)f) (y) = e2πit+πiξ·x−2πiξ·yf(y − x) (23)

defines a continuous function

H × S(R) 3 (g, f)→ ρ(g)f ∈ S(R)

Assuming that, check that (ρ,S(R)) is a representation of H on the linear topologicalspace S(R), i.e. show that ρ(gg′) = ρ(g)ρ(g′). Also show that

EndH(S(R)) = CI .One may check, but it could take too much of your time, that the representation (ρ,S(R))is irreducible and that the same formula (23) defines an irreducible unitary representationof H on L2(R).

Here is one more version of Schur’s Lemma. For a proof see [Wal88, 1.2.2].

Theorem 41. Let (π,V) be an irreducible unitary representation of G. Let V0 ⊆ V be adense G-invariant subspace and let A : V0 → V be a linear G-intertwining map.

Suppose V1 ⊆ V be a dense subspace and let B : V1 → V a linear map such that

(Av0, v1) = (v0, Bv1) (v0 ∈ V0, v1 ∈ V1) .

Then A is a scallar multiple of the identity restricted to V0.

9. Haar measures and extension of a representation of G to a representationof L1(G)

A proof of the following theorem may be found in [HR63, (15.5)-(15.11)]

Theorem 42. There is a positive Borel measure dx on G such that∫G

f(gx) dx =

∫G

f(x) dx (g ∈ G, f ∈ Cc(G)) .

This measure is unique up to a constant multiple. Furthermore there is a group homom-rphism ∆ : G→ R+ such that∫

G

f(xg) dx = ∆(g)

∫G

f(x) dx (g ∈ G, f ∈ Cc(G)) .

Also, the converse of this theorem holds: if a topological group has a left invariantBorel measure the it is locally compact, see[?].

The measure dx is called the left invariant Haar measure on G. The group G is calledunimodular if ∆ = 1. This is certainly the case if G is compact. The Lebesgue measureswe used in sections 2 and 4 are Haar measures and the groups are unimodular. Anexample of a non-unimodular group is

P =

{p =

(a b0 a−1

)a ∈ R×, b ∈ R

}.

Here |a|−2da db is the left invariant Haar measure and ∆(p) = a2.


The convolution φ ∗ ψ of two functions φ, ψ ∈ L1(G) is defined by

φ ∗ ψ(y) =

∫G

φ(x)ψ(x−1y) dx .

The Banach space L1(G) with the product defined by the convolution is a Banach algebra.This algebras may have no identity, but they always has an approximate identity.

Theorem 43. There is a sequence φn ∈ Cc(G) such that

(1) φn ≥ 0 (n = 0, 1, 2, ...) ,

(2)

∫G

φn(x) dx = 1 (n = 0, 1, 2, ...) ,

(3) Given a neighborhood U of the identity in G ,

the support of φn is contained in U for all n sufficiently large.

(See [Lan85, secion 1.1] for a proof.) Given a representation (π,V) the formula

π(φ) =

∫G

φ(x)π(x) (φ ∈ L1(G))

defines a representation of the Banach algebra L1(G) on V, i.e.

π(φ ∗ ψ) = π(φ)π(ψ) (φ, ψ ∈ L1(G)) .

By restriction, (π,V) is also a representation of the convolution algebra Cc(G). Asstraightforward consequence of Theorem 43 we see that a subspace W ⊆ V is G-invariantif and only if it is Cc(G)-invariant. Also, (π,V) is G-irreducible if and only if it is Cc(G)-irreducible.

10. Representations of a compact group

Let G be a compact group with the Haar measure of total mass equal to 1. If (π,V) isa representation of G on a Hilbert space V, then, by averaging the norm on V over G, weobtain another norm with respect to which the representation is unitary.

Hence, while considering representations of G on Hilbert spaces, we may assume thatthey are unitary.

Theorem 44. Any irreducible unitary representation (π,V) of G is finite dimensional.

Proof. Let v ∈ V be a unit vector and let P be the orthogonal projection on the one-dimensional space Cv. Let Q be the continuous linear map defined by

Q =

∫G

π(x)−1Pπ(x) dx .

Then Q = Q∗ commutes with the action of G. Hence Schur’s Lemma, Theorem 36, impliesthat there is a constant c ∈ R such that Q = cI. Since π is unitary,

(Qv, v) =

∫G

(Pπ(x)v, π(x)v) dx =

∫G

(π(x)v, v)(v, π(x)v) dx =

∫G

|(π(x)v, v)|2 dx > 0 .

32 TOMASZ PRZEBINDA

Hence c > 0.Let v1, v2, ... be an orthonormal basis of V. Then for each x ∈ G, π(x)v1, π(x)v2, ...is

also an orthonormal basis of V. Hence∞∑n=1

(Pπ(x)vn, π(x)vn) =∞∑n=1

((π(x)vn, v)v, π(x)vn) =∞∑n=1

|(π(x)vn, v)|2 =‖ v ‖= 1 .

But (Pπ(x)vn, π(x)vn) = (π(x)−1Pπ(x)vn, vn). Hence, after integration over G we seethat

∞∑n=1

(Qvn, vn) = 1 .

Thus c times the number of the elements in the basis is equal to 1. Hence, dimV <∞. �

By combining Theorem 44 with the Spectral Theorem from Linear Algebra we deducethe following corollary.

Corollary 45. An irreducible unitary representation of a compact abelian group is onedimensional.

If dimV < ∞, we let Vc denote the vector space dual to V. The contragredient repre-sentation (πc,Vc) is defined by

πc(g)vc(v) = vc(π(g−1)v) (v ∈ V, vc ∈ Vc, g ∈ G).

Given two finite dimensional representations (π,V) and (π′,V′) define their direct sum(π ⊕ π′,V ⊕ V′) by

(π ⊕ π′)(g)(v, v′) = (π(g)(v), π′(g)(v′)) (g ∈ G, v ∈ V, v′ ∈ V′)

and the tensor product (π ⊗ π′,V ⊗ V′) by

(π ⊗ π′)(g)[v ⊗ v′] = [π(g)(v)]⊗ [π′(g)(v′)] (g ∈ G, v ∈ V, v′ ∈ V′).

By definition the character ΘV = Θπ of a finite dimensional representation (π,V) is thefollowing complex valued function on the group:

ΘV(g) = tr(π(g)) (g ∈ G).

This function is invariant under conjugation

Θ(hgh−1) = Θ(g) (h, g ∈ G).

Also, we have

ΘV⊕V′ = ΘV + ΘV′ , ΘV⊗V′ = ΘVΘV′ and Θπc = Θπ. (24)

Denote by L2(G)G ⊆ L2(G) the subspace of the functions invariant by the conjugation byall the elements of G. Our characters live in L2(G)G.

Lemma 46. Suppose (π,V) is a non-trivial irreducible unitary representation of G. Then∫G

π(x)v dx = 0 (v ∈ V).


Proof. Notice that the integral defines a G-invariant vector u ∈ V. Since π is irreducible,either u = 0 or V = Cu. Since π is non-trivial, the second option is impossible. �

Corollary 47. Suppose (π,V) is a non-trivial irreducible unitary representation of G. Then∫G

π(x) dx = 0 .

Corollary 48. Suppose (π,V) is a finite dimensional representation of G. Then∫G

Θπ(x) dx = dimVG,

where VG ⊆ V is the space of the G-invariant vectors.

Proposition 49. The characters of irreducible representations form an orthonormal setin L2(G)G.

Proof. Consider two such representations (π,V) and (π′,V′). The group G acts on thisvector space HomG(V,V′) by

gT (v) = π′(g)Tπ(g−1)v (g ∈ G, T ∈ HomG(V,V′), v ∈ V).

This way HomG(V,V′) becomes a representation of G. It is easy to check that as such itis isomorphic to (πc ⊗ π′,Vc ⊗ V′). Hence, by (24),

ΘHom(V,V′)(g) = Θπ(g)Θπ′(g) (g ∈ G).

Therefore,

(Θ′π,Θπ) =

∫G

ΘHom(V,V′)(g) dg

= dimension of the space of the G-invariants in HomG(V,V′).

Thus the formula follows from Corollary 39. �

Proposition 50. Any finite dimensional representation of G decomposes into the directsum of irreducible representations.

Proof. This follows from the fact that the orthogonal complement of a G-invariant sub-space is G-invariant. �

Let (π,V) be a finite dimensional representation of G and let

V = V1 ⊕ V2 ⊕ ...⊕ Vn

be the decomposition into irreducibles. The number of the Vj which are isomorphic toV1 is called the multiplicity of V1 in V, denoted m1. We may collect the isomorphicrepresentations in the above formula and (after changing the indecies appropriately) getthe following decomposition

V = V⊕m11 ⊕ V⊕m2

2 ⊕ ...⊕ V⊕mkk ,

34 TOMASZ PRZEBINDA

where the Vj are irreducible and mutually non-isomorphic and

V⊕mjj = Vj ⊕ Vj ⊕ ...⊕ Vj (mj summands).

This is the primary decomposition of V, which is also denoted by

V = m1V1 ⊕m2V2 ⊕ ...⊕mkVk = m1 · V1 ⊕m2 · V2 ⊕ ...⊕mk · Vk,for brevity.

Proposition 51. LetV = m1V1 ⊕m2V2 ⊕ ...⊕mkVk,

be the primary decomposition of V. Then

ΘV = m1ΘV1 +m2ΘV2 + ..mkΘVk

andmj = (ΘV,ΘVj).

In particular V is irreducible if and only if (ΘV,ΘV) = 1.

Proof. This follows from (24) and from Lemma 49. �

Given a finite dimensional representation (π,V) define a map

Mπ : End(V)→ L2(G), Mπ(T )(g) = tr(π(g)T ) = tr(Tπ(g)).

In particular, if I ∈ End(V) is the identity, then

Mπ(I) = Θπ.

Also, it is easy to see that the subspace Mπ(End(V)) ⊆ L2(G) depends only on theequivalence class of π.

Theorem 52. Suppose (π,V) and (π′,V′) are two irreducible unitary representations ofG. Then

a) if (π,V) is not equivalent to (π′,V′), then Mπ(End(V)) ⊥Mπ′(End(V′)),b) for any S, T ∈ End(V), (Mπ(S),Mπ(T )) = 1

dimVtr(ST ∗).

Proof. Define a representation (Π,End(V)) of the group G×G on the vetor space End(V)by

Π(g1, g2)T = π(g2)Tπ(g−11 ) (g1, g2 ∈ G, T ∈ End(V)).

Notice that (Π,End(V)) is isomorphic to outer tensor product (πc ⊗ π,Vc ⊗ V). Hence,∫G×G

|ΘΠ(x1, x2)|2 dx1 dx2 =

∫G×G

|Θπ(x1)Θπ(x2)|2 dx1 dx2

=

∫G

|Θπ(x1)|2 dx1

∫G

|Θπ(x2)|2 dx2 = 1.

Hence, by Proposition 51, (Π,End(V)) is irreducible.We view End(V) as a Hilbert space with the following scalar product

(S1, S2) = tr(S1S∗2) (S1, S2 ∈ End(V)),


and similarly for V′. In particular, we have the adjoint map

M∗π′ : L2(G)→ End(V).

Notice also thatM∗

π′Mπ : End(V)→ End(V′)

is a G×G-intertwining map. If the two representations (π,V) and (π′,V′) are not isomor-phic then The G×G-modules End(V) and End(V′) are not isomorphic. Hence Corollary39 shows that M∗

π′Mπ = 0. Thus for S ∈ End(V) and T ∈ End(V′),

(Mπ(S),Mπ′(T )) = (S,M∗π′Mπ(T )) = 0

and a) follows.Similarly, there is λ ∈ C such that M∗

πMπ = λI. Furtheremore,

λ · dimV = (I,M∗πMπ(I)) = (Mπ(I),Mπ(I)) = (Θπ,Θπ) = 1

and b) follows. �

The following statement is known as the Peter-Weyl Theorem, see [PW27].

Theorem 53. There are the following direct sum orthogonal decompositions of Hilbertspaces

a) L2(G) =∑

π∈G Mπ(End(V)),b) L2(G)G =

∑π∈G CΘπ.

Proof. Part b) follows from part a), because, by Theorem 38, End(V)G = CI. Indeed,

L2(G)G =∑π∈G

(Mπ(End(V)))G =∑π∈G

Mπ(End(V)G),=∑π∈G

CΘπ.

where only the middle equation

(Mπ(End(V)))G = Mπ(End(V)G)

requires an explanation. We need to show that if

tr(Tπ(ghg−1)) = tr(Tπ(h))

for all h ∈ G, then π(g−1)Tπ(g) = T . Since tr(Tπ(ghg−1) = tr(π(g−1)Tπ(g)π(h)), we’llbe done as soon as we show that if tr(Sπ(h)) = 0 for all h ∈ G, then S = 0. But, sincethe map Mπ is injective (because π is irreducible), this is indeed the case.

Part a) requires some work. We follow the argument in [Kna86, Theorem 1.12]. LetU =

∑π∈G Mπ(End(V)). This is a subspace of L2(G) closed under the left and right trans-

lations and under the ∗ operation, φ → φ∗, φ∗(x) = φ(x−1). Hence so is the orthogonalcomplement U⊥ ⊆ L2(G).

Suppose U⊥ 6= {0}. We shall arrive at a contradiction. Let φ ∈ U⊥ be non-zero.Let φn ∈ Cc(G) be an approximate identity, as in Theorem 43. Then a straightforwardargument shows that

limn→∞

‖ φn ∗ φ− φ ‖2= 0 .

36 TOMASZ PRZEBINDA

Hence there is n such that φn ∗ φ 6= 0. But this function is continuous. Thus we mayassume that φ is continuous. Furthermore, applying the translations and the ∗ operationwe may assume that φ(1) > 0 and φ = φ∗. Replacing φ by the integral∫

G

L(x)R(x)φ dx

we may assume that φ is invariant under conjugation. Let

Tψ(x) =

∫G

φ(x−1y)ψ(y) dy .

Since the integral kernel φ(x−1y) is continuous on G × G, T is a compact operator onL2(G). Furthermore, T = T ∗ 6= 0, because φ = φ∗. Hence, by Theorem 32, T has anon-zero finite dimensional eigenspace Vλ ⊆ L2(G). Since T commutes with L(G), Vλis closed under L(G). Hence Theorem 50 shows that there is a G-irreducible subspaceWλ ⊆ Vλ. Let fj be an orthonormal basis of Wλ. Set

hi,j(x) = (L(x)fi, fj) =

∫G

fi(x−1y)fj(y) dy .

This is a matrix coefficient of an irreducible representation of G, thus it belongs to U.Therefore,

0 = (φ, hi,i) =

∫G

∫G

φ(x)fi(x−1y)fi(y) dy dx

=

∫G

∫G

φ(yx−1)fi(x)fi(y) dy dx =

∫G

∫G

φ(x−1y)fi(y) dy fi(x) dx

=

∫G

Tfi(x) fi(x) dx = λ

∫G

fi(x)fi(y) dx ,

which is a contradiction. �

Theorem 54. Any unitary representation (ρ,W) of G is completely reducible, i.e. theHilbert space W is the orthogonal sum of irreducible finite dimensional representations ofG.

Proof. We follow [Kna86, Theorem 1.12]. Suppose (ρ,W) is not completely reducible. ByZorn’s Lemma we may choose a maximal orthogonal set of finite dimensional irreducibleinvariant subspaces. Let U denote the closure of their sum. Suppose 0 6= v ∈ U⊥. We’llarrive at a contradiction.

Let φn ∈ Cc(G) be an approximate identity, as in Theorem 43. Then, as we have seenbefore, there is n such that ρ(φn)v 6= 0. We fix this n.

Theorem 53 a) implies that there is a finite set F ⊆ G and φ ∈⊕

π∈F Mπ(End(V)),such that

‖ φn − φ ‖2≤1

2 ‖ v ‖‖ ρ(φn)v ‖ .


Since the total mass of G is 1, we have

‖ φn − φ ‖1≤‖ φn − φ ‖2 .

Hence

‖ ρ(φn)v − ρ(φ)v ‖≤‖ φn − φ ‖1‖ v ‖≤1

2‖ ρ(φn)v ‖ .

Therefore

‖ ρ(φ)v ‖≥‖ ρ(φn)v ‖ − ‖ ρ(φn)v − ρ(φ)v ‖≥ 1

2‖ ρ(φn)v ‖> 0 .

This is a contradiction because ρ(φ)v lies in a finite dimensional invariant subspace ofW. �

For a function φ ∈ L1(G) define the Fourier transform

Fφ(π) =

∫G

φ(x)π(x) dx = π(φ) . (25)

Notice that in order to be consistent with the theory of Fourier series, section 4, we shouldreplace π by πc in this definition. However we shall follow the tradition and not do that.

Thus for a representation (π,V), Fφ(π) ∈ End(V). Notice that

π(g)Fφ(π) = F(L(g)φ) (g ∈ G) . (26)

Set d(π) = dimV) = Θπ(1). This is the degree of the representation (π,V).

Theorem 55. (Fourier inversion for G) For any φ ∈⊕

π∈G Mπ(End(V)) (algebraic sum),

φ(g) =∑π∈G

d(π) · tr(Fφ(π)π(g−1)) (g ∈ G) , (27)

or equivalently

φ(1) =∑π∈G

d(π) ·Θπ(φ) (g ∈ G) , (28)

Notice that in the case of the classical Fourier series, section 4, the above formula (27)refers only to trigonometric polynomials, φ.

Proof. Clearly (28) is a particular case of (27). Also, (27) follows from (28) and (26).Let us write (π,Vπ) for (π,V) in order to indicate the dependence of the vector space

on π. By definition, there is a finite set F ⊆ G and operators Tπ ∈ End(Vπ) such that

φ(g) =∑ρ∈F

tr(Tρρ(g)) (g ∈ G) .

Hence

Fφ(π) =∑ρ∈F

∫G

tr(Tρρ(g))π(g) dg .

38 TOMASZ PRZEBINDA

Therefore, by Theorem 52

trFφ(π) =∑ρ∈F

∫G

tr(Tρρ(g)) trπ(g) dg =

∫G

tr(Tππ(g)) trπ(g) dg

= (Tππ(g)), π(g)) =1

dπtr(Tππ(g)π(g)∗) =

1

dπtr(Tπ) .

Hence ∑π∈G

d(π) ·Θπ(φ) =∑π∈G

d(π) · trFφ(π) =∑π∈G

tr(Tπ) = φ(1).

�

Theorem 56. (Parseval formula) For any φ ∈⊕

π∈G Mπ(End(V)),

‖ φ ‖22=∑π∈G

dπ ‖ Fφ(π) ‖2 .

Proof. That the left hand side is equal to

φ ∗ φ∗(1) =∑π∈G

dπc tr(Fφ(π)Fφ(π)∗),

which coincides with the right hand side. �

10.1. Unique factorization domains and Hilbert’s Nulstellensatz. This section isincluded to explain Hilbert’s Nullstellensatz, to be used in the following section.

A ring R, with an identity, is called a Unique Factorization Domain (UFD) if and only ifR has no zero divisors and every element ofR decomposes into a product of indecomposibleelements uniquelly up a permutation of factors and multiplication by units. In particular,any finite subset of an UFD has the greatest common divisor, and any indecomposableelement of an UFD is prime.

Proposition 57. Let F be a field. Then the ring F [X] is a UFD.

Proof. Euclid’s algorithm shows that R = F [X] is a Principal Ideal Domain. Hence, ifa, b ∈ R, then the ideal generated by a and b,

(a, b) = (c)

for some c ∈ R.In fact c is a greatest common divisor of a and b, denoted gcd(a, b). Indeed, since

a ∈ (c) and b ∈ (c), we see that c divides a and b. Hence c divides gcd(a, b). On the otherhand,

c = a′a+ b′b

for some a′, b′ ∈ R. Thus gcd(a, b) divides c.If p ∈ R is indecomposable, then p is prime. Indeed, suppose a, b ∈ R are such that

p divides ab but does not divide a. Since p is indecomposable we see that gcd(p, a) = 1.Hence, there are a′ ∈ R and p′ ∈ R such that

1 = a′a+ p′p.


Thereforeb = a′(ab) + p′pb.

Thus p divides b.By looking at the degrees of the polynomials we see that any element a ∈ R decomposes

into a product of indecomposables. Suppose we have two such decompositions

a = p1p2...pr = q1q2...qs.

Since p1 is prime, it divides one of the qi. We may assume that it divides q1. Hence thereis a unit u1 such that q1 = p1u1. Since R has no zero divisors we see that

p2...pr = u1q2...qs.

Repeating this process we see that the decomposition is unique, as stated. �

Theorem 58. If R is a UFD then R[x] is a UFD.

Proof. A polynomial in R[x] is called primitive if the greatest common divisor of itscoefficients is 1. In particular any primitive element of R[x] is indecomposable if andonly if it cannot be written as the product of two elements of positive degree. Also, anyelement of R[x] is the product of an element of R and a primitive polynomial.

Also, it is clear that any indecomposable element of positive degree in R[x] is theproduct of a prime in R and an indecomposable primitive element of the same degree.

Since any element of R[x] is the product of indecomposables we’ll be done as soon aswe show that the decomposition is unique (up to a permutation and multiplication byunits). Let F be the field of fractions of R. Then

R[x] ⊆ F [x].

Hence, by Proposition 57, the only fact we need to verify is that a primitive indecompos-able element f ∈ R[x] of positive degree is also indecomposable in F [x].

Suppose, there are gF , hF ∈ F [x], of positive degree, such that

f = gFhF .

There is b ∈ R such that bgF ∈ R[x]. Also, there is a ∈ R and a primitive g ∈ R[x], ofthe same degree as gF , such that

bgF = ag.

Hence,

f = g(abhF

).

There is c ∈ R such that

c(abhF

)∈ R[x].

If c is a unit in R then we see that f decomposes in R[x], a contradiction.Suppose c is not a unit. We see that g divides cf in R[x]. Take c with minimal possible

number of primes in it such that g divides cf in R[x]. Thus there is h ∈ R[x] such that

cf = gh.

40 TOMASZ PRZEBINDA

Let p ∈ R be a prime which divides c.Since p divides the left hand side, it divides the right hand side. Since g is primitive, p

divides h in R[x]. Thus

c

pf = g

(1

ph

),

where(

1ph)∈ R[x], a contradiction. �

Now we turn to Hilbert’s Nullstellensatz. We present an argument made by DanielAllcock, which according to him goes back to Zariski.

Theorem 59. Let k be a field and K a field extension which is finitely generated as ak-algebra. Then K is algebraic over k.

Proof. We will assume throughout that K is transcendental over k and finitely generatedas a k-algebra, and deduce that K is not finitely generated as a k-algebra, a contradiction.

Suppose first thatK has transcendence degree one; this means that it contains a subfieldk(x) which is a copy of the one-variable rational function field, and that K is algebraic overk(x). This, together with the finite generation of K, shows that K has finite dimensionas a k(x)-vector space. Choose a basis e1, ..., el and write down the multiplication tablefor K:

eiej =∑k

aijk(x)

bijk(x)ek,

with the a’s and b’s in k[x]. We will show that for any f1, ..., fm ∈ K, the k-algebra Athey generate is smaller than K.

It is convenient to adjoin f0 = 1 as a generator. Express f0, ..., fm in terms of our basis:

fi =∑j

cij(x)

dij(x)ej,

with the c’s and d’s in k[x]. Now, an element a of A is a k-linear combination of f0 = 1and products of f1, ..., fm. Expanding in terms of our basis, we see that a is a k(x)-linearcombination of products of the ei, with the special property that the denominators of thecoefficients involve only the d’s. Using the multiplication table repeatedly, we see that ais a k(x)-linear combination of the ei, whose coefficients’ denominators involve only theb’s and d’s.

A precise way to state the result of this argument is: when a is expressed as a k(x)-linear combination of the ei, with every coefficient in lowest terms, then all its coefficients’denominators’ irreducible factors are among the irreducible factors of the b’s and d’s.Therefore

1

some other irreducible polynomial

cannot lie in A, and A is smaller than K.


This argument requires the existence of infinitely many irreducible polynomials in k[x];to prove this one can mimic Euclid’s proof of the infinitude of primes in Z. (If k is infinitethen one can just take the infinitely many linear polynomials x− c, c ∈ k.)

Now suppose K has transcendence degree > 1 over k, and choose a subextension Fover which K has transcendence degree 1. By the above, K is not finitely generated asa F -algebra, so it isn’t as a k-algebra either. (To build F explicitly, choose k-algebragenerators x1, ..., xn for K over k and set F = k(x1, ..., xl−1), where xl is the last of thex’s which is transcendental over the field generated by its predecessors.) �

Theorem 60. (The ‘Weak’ Nullstellensatz’.)Let k be an algebraically closed field. Then every maximal ideal in the polynomial

ring R = k[x1, ..., xn] has the form (x1 − a1, ..., xn − an) for some a1, ..., an ∈ k. As aconsequence, a family of polynomial functions on kn with no common zeros generates theunit ideal of R, i.e. R.

Proof. If m is a maximal ideal of R then R/m is a field which is finitely generated as ak-algebra. By the previous theorem it is an algebraic extension of k, hence equal to k.Therefore each xi maps to some ai ∈ k under the natural map R → R/m = k, so mcontains the ideal (x1 − a1, ..., xn − an). This is a maximal ideal, so it equals m.

To see the second statement, consider the ideal generated by some given polynomialfunctions with no common zeros. If it lay in some maximal ideal, say (x1−a1, ..., xn−an),then all the functions would vanish at (a1, ..., an) ∈ kn, contrary to hypothesis. Since itdoesn’t lie in any maximal ideal, it must be all of R. �

Theorem 61. (Nullstellensatz.)Suppose k is an algebraically closed field and g and f1, ..., fm are members of R =

k[x1, ..., xn], regarded as polynomial functions on kn. If g vanishes on the common zero-locus of the fi, then some power of g lies in the ideal they generate.

Proof. The polynomials f1, ..., fm and xn+1g− 1 have no common zeros in kn+1, so by theweak Nullstellensatz we can write 1 = p1f1 + +pmfm+pm+1 ·(xn+1g−1) ,where the p’s arepolynomials in x1, ..., xn+1. Taking the image of this equation under the homomorphismk[x1, ..., xn+1] → k(x1, ..., xn) given by xn+1 → 1/g, we find 1 = p1(x1, ..., xn, 1/g)f1 + +pm(x1, ..., xn, 1/g)fm. After multiplying through by a power of g to clear denominators,we have Hilbert’s theorem. �

10.2. Spherical harmonics in Rn. The Laplacian in Rn, n ≥ 2, may be written as

∆n = ∂2x1

+ ∂2x2

+ ...+ ∂2xn = ∂2

r +n− 1

r∂r +

1

r2L, (29)

where r2 = x21 + x2

2 + ...+ x2n and L is the Laplacian on the sphere Sn−1.

Lemma 62. If φ is a smooth harmonic function on Rn \ {0}, homogeneous of degree k,then Lφ = −k(k + n− 2)φ.

Proof. The assumption that φ is harmonic means that ∆nφ = 0. Thus (29) shows that

0 = (∂2r +

n− 1

r∂r +

1

r2L)φ.

42 TOMASZ PRZEBINDA

Therefore

Lφ = −r2∂2rφ− (n− 1)r∂rφ = −((r∂r)

2 − r∂r)φ− (n− 1)r∂rφ

= −(r∂r)2φ− (n− 2)r∂rφ = −k2φ− (n− 2)kφ.

�

Lemma 63. Let P denote the space of the complex valued polynomial functions on Rn.For φ, ψ ∈ P define

(φ, ψ) = φ(∂x1 , ∂x2 , ..., ∂xn)ψ(0).

Then ( , ) is a positive definite hermitian form on P. Moreover

(φ, ψη) = (∂(ψ)φ, η) (φ, ψ, η ∈ P).

Proof. In the multi-index notation, let

φ(x) =∑α

aαxα, ψ(x) =

∑β

bβxβ.

Then(φ, ψ) =

∑α,β

aαbβ∂αxβ|x=0 =

∑α

aαbα α!.

Hence the form ( , ) is hermitian and positive definite. Furthermore,

(φ, ψη) = (ψη, φ) = ∂(ψ)∂(η)φ(0)

∂(η)∂(ψ)φ(0) = ∂(η)∂(ψ)φ(0) = (η, ∂(ψ)φ) = (∂(ψ)φ, η).

�

Lemma 64. Let Pk ⊆ P be the subspace of the polynomials of degree k and let Hk ⊆ Pkbe the subspace of the harmonic polynomials. Then we have the following direct sumorthogonal decomposition:

Pk = Hk + r2Hk−2 + r4Hk−4 + ...+ r2mHk−2m,

where m is the largest integer such that 2m ≤ k.

Proof. Let φ ∈ Pk. Notice that φ ∈ Hk iff

(∆nφ, ψ) = 0 for all ψ ∈ Pk−2,

or equivalently, by Lemma 63,

(φ, r2ψ) = 0 for all ψ ∈ Pk−2.

HenceHk = (r2Pk−2)⊥ ∩ Pk.

ThereforePk = Hk + r2Pk−2 = Hk + r2Hk−2 + r4Pk−2 = ...

�


Theorem 65. Explicitly, for k ≥ 2,

Hk = span {(c1x1 + c2x2 + ...+ cnxn)k; c21 + c2

2 + ...+ c2n = 0}.

Proof. It is easy to check that the right hand side of the above equation is contained inthe left hand side.

Consider an element φ ∈ Hk which is orthogonal to the space on the right hand side.Then

0 = (φ, (c1x1 + ...+ cnxn)k) = ∂(φ)(c1x1 + ...+ cnxn)k(0)

= ∂(φ)∑β

k!

β1!...βn!cβx

β|x=0 =∑α

φαcαk! = k!φ(c).

Thus φ vanishes on the variety

V = {c ∈ Cn; c21 + ...+ c2

n = 0}. (30)

Let I = r2P . This is an ideal in P . Recall Hilbert’s Nullstellenstatz (Theorem 61):

Ideal(V ariety(Ideal)) =√Ideal.

Since for our ideal I and our variety V ,

V ariety(I) = V ,we see that φ ∈

√I. Thus there is p = 1, 2, 3... such that

φp ∈ r2P .Suppose n ≥ 3. Then a simple computation using the homogeneity of r2 shows that r2 isan irreducible polynomial. Since P is a Unique Factorization Domain (see Theorem ??),r2 is prime and therefore

φ ∈ r2P .Hence,

φ ∈ r2Pk−2.

Lemma 64 shows that φ = 0. Thus the equality follows.Suppose n = 2. Then Reφ is a harmonic polynomial. Hence there is a holomorphic

function ψ(x1 + ix2), in fact a polynomial in x1 + ix2, such that Reφ(x1, x2) = Reψ(x1 +ix2). Thus

Reφ(x1, x2) =1

2(ψ(x1 + ix2) + ψ(x1 + ix2)),

which is in the desired space. Similarly we deal with the imaginary part. �

Lemma 66. Let Uk = Hk|Sn−1. Then L2(Sn−1) =∑∞

k=0 Uk.

Proof. The algebra P|Sn−1 is closed under conjugation and separates points of the sphere.Hence, the algebraic direct sum of the Uk is dense in C(Sn−1), and therefore also inL2(Sn−1). Furthermore, by Lemma 62, Uk is the −k(n + k − 2)-eigensubspace for theaction of the selfadjoint operator L. Since all these eigenvalues are different the subspacesare orthogonal. �

44 TOMASZ PRZEBINDA

(Notice that the hermitian form on the space of the polynomials defined in Lemma 63is not the same as the one given by the restriction to the sphere and integration there.Indeed, the homogeneous polynomials of different homogeneity degrees are orthogonalwith respect to the hermitian form defined in Lemma 63, but they may have the samerestriction to the unit sphere (r2|Sn−1 = 1).)

Define a representation ρ of the group O(n) on the Hilbert space L2(Sn−1) by theformula

ρ(g)φ(σ) = φ(g−1σ) (g ∈ O(n), σ ∈ Sn−1, φ ∈ L2(Sn−1)). (31)

Theorem 67. For any k = 0, 1, 2, ..., the restriction of ρ to Uk is an irreducible represen-tation of O(n).

Proof. Let en = (0, 0, ..., 1) ∈ Sn−1 be the north pole. Since the evaluation at en is acontinuous linear functional on the finite dimensional space Uk, there is φ ∈ Uk such that

ψ(en) =

∫Sn−1

ψ(σ)φ(σ) dµ(σ) (ψ ∈ Uk).

Let O(n − 1) ⊆ O(n) be the stabilizer of en. Clearly φ is O(n − 1)-invariant. But φ, orrather its k-homogeneous extension to Rn, is harmonic.

Consider an arbitrary O(n − 1)-invariant function η ∈ Hk. Since an On−1-invariantpolynomial on Rn−1 is a polynomial in the variable r2

n−1 = x21 + x2

2 + ...+ x2n−1,

η(x) =m∑j=0

cjxk−2jn r2j

n−1,

where m is the greatest integer ≤ k2. Furthermore,

0 =∑

0≤j≤ k2

cj(∂2xn(xk−2j

n r2jn−1) + xk−2j

n ∆n−1r2jn−1)

=∑

0≤j≤ k−22

cj(k − 2j)(k − 2j + 1)xk−2j−2n r2j

n−1 +∑

1≤j≤ k2

cj2j(n+ 2j − 2)x2k−2jn r2j−2

n−1

=∑

1≤j≤ k2

(cj−1(k − 2j + 2)(k − 2j + 1) + cj2j(n+ 2j − 2))xk−2jn r2j

n−1.

Therefore,

cj = −cj−1(k − 2j + 2)(k − 2j + 1)

2j(n+ 2j − 2)(1 ≤ j ≤ k

2).

We see that the space of the O(n− 1)-invariants in Uk is one-dimensional.Let V ⊆ Uk be a non-zero O(n)-invariant subspace. Then there is ψ ∈ V such that

ψ(en) 6= 0. Let

ψ(σ) =

∫O(n−1)

ψ(gσ) dg.


Then

ψ(en) =

∫O(n−1)

ψ(gen) dg =

∫O(n−1)

ψ(en) dg 6= 0.

Thus ψ is a non-zero O(n− 1)-invariant in Uk. Since ψ ∈ V we see that φ ∈ V. Thereforeφ is in every non-zero invariant subspace of Uk. This shows that Uk is irreducible. �

Lemma 68.

dimPk =

(n+ k − 1n− 1

)(32)

Proof. The dimension is equal to the number of ways one may put k balls in n boxes.Imagine that these boxes are lined up in a row (along the x-axis). In front of each boxline up all the balls that go into it (in the direction of the y-axis). Suppose these ballsare blue. Insert a red ball between two consecutive boxes. We end up with k + (n − 1)balls, n− 1 of them red. The number of such arrangements is (32). �

Corollary 69. The space of the O(n−1)-invariants in Uk is one-dimensional. In particularthere is a unique function φk ∈ Uk such that φk(en) = 1. Furthermore,

dimUk =

(n+ k − 1n− 1

)−(n+ k − 3n− 1

).

11. Square integrable representations of a locally compact group

In this section G is a locally compact group. We follow [Wal88, section 1.3].

Lemma 70. Let (τ,V) be an irreducible unitary representation of G. Suppose that forsome non-zero vectors u0, v0 ∈ V∫

G

|(τ(x)u0, v0)|2 dx <∞ . (33)

Then for arbitrary u, v ∈ V, ∫G

|(τ(x)u, v)|2 dx <∞ . (34)

Moreover, the map T : V→ L2(G) defined by

Tu(x) = (τ(x)u, v0) (u ∈ V, x ∈ G)

is G-intertwining and has the property that there is t > 0 such that

(Tu, Tv) = t(u, v) (u, v ∈ V) . (35)

Proof. We may assume that u0, v0 ∈ V are unit vectors. Let W0 be the linear span of allthe vectors π(g)u0, g ∈ G. This is a G-invariant subspace of V. Since V is irreducible, W0

is dense in V. Let

W = {u ∈ V;

∫G

|(τ(x)u, v0)|2 dx <∞} .

Then W is also a G-invariant subspace of V and W contains W0. Hence W is dense in V.

46 TOMASZ PRZEBINDA

ClearlyTτ(g)u = R(g)Tu (g ∈ G) .

Define an inner product on W by

〈w1, w2〉 = (w1, w2) + (Tw1, Tw2) (w1, w2 ∈ W) .

One checks that W with this inner product is complete. (Every Cauchy sequence in Whas a limit in W.) Thus (R,W) is a unitary representation of G.

The inclusionIW : W 3 w → w ∈ V

is a bounded linear map which intertwines the actions of G. The adjoint

I∗W : V→ W

is also a bounded linear map which intertwines the actions of G. Now we apply Theorem41 with A = I∗W, V0 = V, B = IW and V1 = W. The conclusion is that I∗W is a scalarmultiple of the identity. Hence W = V. Since any two elements u, v ∈ W satisfy (34), theclaim (34) follows.

Now V is equipped with two inner products (·, ·) and 〈·, ·〉 preserved by G. Define amap S : V→ V by

〈u, v〉 = (Su, v) .

Then S commutes with the action of G. Hence S = sI for some s ∈ C. Taking u = v 6= 0we see that s > 0. This implies (35). �

An irreducible unitary representation of G is called square integrable if one, or equiva-lently all, its matrix coefficients are square integrable.

Proposition 71. Let (σ,U) and (τ,V) be irreducible unitary square integrable representa-tions of G. Then∫

G

(σ(x)u1, u2)(v2, τ(x)v1) dx = 0 (u1, u2 ∈ U , v1, v2 ∈ V) if σ 6' τ (36)

There is a positive number d(τ) such that∫G

(τ(x)u1, u2)(v2, τ(x)v1) dx =1

d(τ)(u1, v1)(v2, u2) (u1, u2, v1, v2 ∈ V) . (37)

Proof. DefineSu(x) = (σ(x)u, u2) (u ∈ U, x ∈ G)

andTv(x) = (τ(x)v, v2) (v ∈ V, x ∈ G) .

Then (35) shows that there is a positive constant C such that for u ∈ U and v ∈ V

|(Su, Tv)| ≤‖ Su ‖2‖ Tv ‖2≤ C ‖ u ‖‖ v ‖ .Hence there is a bounded G-intertwinig linear map A : V→ U such that

(Su, Tv) = (u,Av) (u ∈ U, v ∈ V) .


This map is zero if the representations are not equivalent. Since∫G

(σ(x)u1, u2)(v2, τ(x)v1) dx = (Su1, T v1) = (u1, Av1) ,

(36) follows. Suppose σ = τ . In this case let us write Tv2 for T and Tu2 for S. Then

(Tu2u, Tv2v) = (Su, Tv) = (u,Av) (u, v ∈ V) ,

where A is a constant because the representation is irreducible. Thus A = aI for somea ∈ C. This constant a = au2,v2 depends continuously and anti-linearly on u2 and linearlyv2. Thus there is a constant b such that

au2,v2 = b(v2, u2) (u2, v2 ∈ V) .

By taking u2 = v2 we see that b > 0.∫G

(τ(x)u1, u2)(v2, τ(x)v1) dx = au2,v2(u1, v1) = b(u1, v1)(v2, u2) .

�

The constant d(τ) is called the formal degree of τ . It depends on the choice of the Haarmeasure on G. If G is compact with the Haar measure of total mass 1 then, as we haveseen in Theorem 52, d(τ) = dimV. Hence the name.

12. G = SL2(R)

12.1. The finite dimensional unitary representations.

Theorem 72. The only finite dimensional unitary representation of the group SL2(R) isthe trivial representation.

Proof. Let (ρ,V) be a finite dimensional representation of the group SL2(R). We shallwork under the additional assumption that the group homomorphism

SL2(R) 3 g → ρ(g) ∈ GL(V)

is a polynomial map. We shall see later in these notes that this assumption is not neces-sary. Then the above map extends to a group homomorphism

SL2(C) 3 g → ρ(g) ∈ GL(V) .

In particular for any X ∈ sl2(C) we have

dρ(X)v =d

dtρ(exp(tX))v|t=0 .

This way we obtain a C-linear map

dρ : sl2(C)→ End(V) .

Since the subgroup SU2 ⊆ SL2(C) is compact, there exists a positive definite hermitianform (·, ·) on V preserved by SU2. Hence all the operators

dρ(X) , X ∈ su2

48 TOMASZ PRZEBINDA

are skew-hermitian and therefore have imaginary eigenvalues.On the other hand, suppose there exists an SL2(R)-invariant positive definite hermitian

form 〈·, ·〉 on V. Then all the operators

dρ(X) , X ∈ sl2(R)

are skew-hermitian and therefore have imaginary eigenvalues. Define

p = isu2 ∩ sl2(R) .

Then the operatorsdρ(X) , X ∈ p

are skew-hermitian and therefore have imaginary eigenvalues. But dρ(iX) = idρ(X).Hence the operators

dρ(X) , X ∈ p

are both hermitian and skew-hermitian. Thus

dρ(p) = {0} .It is easy to check that the commutator

[p, p] = so2 .

Butsl2(R) = so2 + p .

Thereforedρ(sl2(R)) = {0} .

Henceρ(exp(sl2(R))) = {I} .

But since the group SL2(R) is connected it is generated by exp(sl2(R)). Thus the repre-sentation is trivial. �

This is a particular case of a general theorem of Segal and von Neumann, [SvN50].Define the following subgroups of G:

K =

{(cos θ sin θ− sin θ cos θ

); θ ∈ R

},

A =

{(a 00 a−1

), a > 0

},

N =

{(1 n0 1

), n ∈ R

}.

The following map, is a bijective diffeomorphism

A× N×K 3 (a, n, k)→ ank ∈ G . (38)

In particular G = ANK. This is called the Iwasawa decomposition of G.

Problem 14. Prove that the map (38) is bijective.


This is straightforward.Let S be the subset of G consisting of the symmetric and positive definite matrices.

Then

G = KS . (39)

This is the polar decomposition of G.

Problem 15. Prove that the map

K× S 3 (k, s)→ ks ∈ G (40)

is a surjective.

This follows from Spectral Theorem.

Problem 16. Prove that the map

K× A×K 3 (k, n, k)→ kak ∈ G (41)

is a surjective. In particular G = KAK. This is called the Cartan decomposition of G.

This is immediate from (40) and Spectral Theorem.

12.2. The maximal compact subgroup K = SO2(R).We let

k(θ) =

(cos θ sin θ− sin θ cos θ

)and define

K = SO2(R) = {k(θ); θ ∈ R} .This is a maximal compact subgroup of G, which is unique up to conjugation. The groupK is commutative. Define the characters

χn(k(θ)) = einθ (θ ∈ R, n ∈ Z) .

For two integers n and m define

Sn,m = {f ∈ Cc(G); f(k(θ1)gk(θ2) = e−inθ1f(g)e−imθ2} .

Lemma 73. The algebraic sum⊕

n,m Sn,m is L1-dense in Cc(G). In fact, given ε > 0 and

f ∈ Cc(G), there exists a function g ∈⊕

n,m Sn,m such that the support of g is contained

in K(supp f)K, and such that ‖ f − g ‖∞< ε..

Proof. This follows from the properties of the Fourier series. See [Lan85, page 20]. �

Lemma 74. The following formulas hold.

Sn,m ∗ Sl,q = {0} if m 6= l ,

S∗n,m = Sm,n ,

Sn,m ∗ Sm,q ⊆ Sn,q .

Proof. This is straightforward [Lan85, page 22]. �

50 TOMASZ PRZEBINDA

Lemma 75. Let S = {x ∈ G; x = xt}. Then the map

K× S 3 (k, s)→ ks ∈ G

is a bijective diffeomrophism.

Proof. This is well known from Linear Algebra. See straightforward [Lan85, page 22]. �

Lemma 76. The algebra S0,0 is commutative.

Proof. The argument is due to I. M. Gelfand. Consider an element x ∈ G. Then there isk ∈ K and s ∈ S such that x = ks. Hence the transpose

xt = ktst = k−1s = k−1xk−1 . (42)

Also, the group G is unimodular. Hence the Haar measure is invariant under the changeof variables x→ xt. Let f, g ∈ S0,0. Set f t(x) = f(xt). It is easy to check that

(f ∗ g)t = gt ∗ f t .Notice that f t = f . Indeed,

f t(x) = f(xt) = f(k−1xk−1) = f(x) ,

because elements of S0,0 are K-bi-invariant. Hence

gt ∗ f t = g ∗ f .By Lemma 74, f ∗ g ∈∈ S0,0. Hence (f ∗ g)t = f ∗ g. The conclusion follows. �

Lemma 77. For any n ∈ Z, the algebra Sn,n is commutative.

Proof. The argument is due to S. Lang, see [Lan85]. Let

γ =

(1 00 −1

).

Clearly γ = γ−1. Let s = st ∈ G. Then there is k ∈ K such that ksk−1 = d is diagonal.Therefore

γsγ = γk−1dkγ = γk−1γγdγγkγ = γk−1γdγkγ .

Thus there is ks ∈ K such thatγsγ = k−1

s sks .

Notice thatγkγ = k−1 (k ∈ K) .

Recall that f t(x) = f(xt) and let fγ(x) = f(γxγ). Then, for f ∈ Sn,nfγ(x) = f t(x)

Indeed, write x = ks. Then

fγ(ks) = f(γksγ) = f(k−1γsγ) = χn(k)f(γsγ)

= χn(k)f(k−1s sks) = χn(k)χn(ks)f(s)χn(ks)

−1

= χn(k)f(s) = f(s)χn(k) = f(sk−1) = f((ks)t)

= f t(ks) .


Therefore for any f, g ∈ Sn,n,

fγ ∗ gγ = f t ∗ gt = (g ∗ f)t = (g ∗ f)γ = gγ ∗ fγ

and we are done. �

For a representation (π,V) of G on a Banach space V define

Vn = {v ∈ V; π(k(θ))v = einθv, θ ∈ R} .This is the χn isotypic component of V.

Lemma 78. Let (π,V) be a representation of G on a Banach space V. Then

π(Sn,m)V ⊆ Vn

and for q 6= m,π(Sn,m)Vq = 0 .

Proof. See [Lan85, page 23]. �

Lemma 79. Let (π,V) be an irreducible representation of G on a Banach space V. Thenfor any q such that Vq 6= 0, the space Vq is Sq,q-irreducible. Also, π(Sq,q)Vq 6= 0.

Proof. This follows from the fact that the ∗ algebra⊕n,m

Sn,m

is L1-dense in Cc(G) see [Lan85, page 24]. �

Theorem 80. Let (π,V) be an irreducible unitary representation of G on a Hilbert spaceV. Fix an integer n. Then dimVn = 1 or 0.

Proof. In this case π(Sn,n) is a commutative ∗ algebra. Hence the claim follows fromTheorem 34. �

From now on we consider only the representations for which dimVn ≤ 1 for all n ∈ Z.

Theorem 81. Let (π,V) be an irreducible representation of G on a Banach space V. Thenthe space of finite sums ⊕

n∈Z

Vn ⊆ V

is dense.If (π,V) be an irreducible unitary representation of G on a Hilbert space space V, then

V =∑n∈Z

Vn

is a Hilbert space direct sum orthogonal decomposition.

Proof. This follows from the fact that the ∗ algebra⊕n,m

Sn,m

is dense in Cc(G) see [Lan85, page 25]. �

52 TOMASZ PRZEBINDA

12.3. Induced representations. Let us fix Haar measures on these groups

d

(cos θ sin θ− sin θ cos θ

)=

1

2πdθ , d

(a 00 a−1

)=da

a, d

(1 n0 1

)= dn .

Then, in terms of the Iwasawa decomposition, the formula

dx = da dn dk (x = ank, a ∈ A, n ∈ N, k ∈ K)

defines a Haar measure on G. Furthermore

α

(a 00 a−1

)= a2 , ρ

(a 00 a−1

)= a .

Let P = AN. This is a subgroup of G with the modular function

∆

((a 00 a−1

)(1 n0 1

))= a2 .

Theorem 82. Let (σ,V) be a representation of P on a Hilbert space V. Denote by V(σ)the space of functions f : G→ V such that

f(px) = ∆(p)12σ(p)f(x) (x ∈ G)

whose restriction to K is square integrable:∫K

‖ f(k) ‖2 dk <∞ .

Setπ(g)f(x) = f(xg) (g, x ∈ G)

Then (π,V(σ)) is a representation of G. If σ is unitary then so is π.

Proof. See [Lan85, Theorem 2, page 44]. �

Our main example is going to be the case when V = C and for some fixed complexnumber s,

σ(an) = ρ(a)s (a ∈ A, n ∈ N) .

Then we shall write V(s) = V(σ) and πs = π. Then the transformation property offunctions in V(s) looks as follows

f(anx) = ρ(a)1+sf(x) (a ∈ A, n ∈ N, x ∈ G) . (43)

The resulting representation(πs,V(s)) is called the principal series representation.

Lemma 83. Letρs(ank) = ρ(a)1+s (a ∈ A, n ∈ N, k ∈ K) .

Then ρs ∈ V(s) is K-ivariant, has norm 1, and

(πs(x)ρs, ρs) =

∫K

ρs(kx) dk =

∫K

ρ(kx)1+s dk .

Proof. See [Lan85, page 47]. �


The function

φs =

∫K

ρ(kx)1+s dk (44)

is known as a spherical function.

Lemma 84. For a function ψ ∈ Cc(G), πs(ψ) is an integral kernel operator with theintegral kernel

qψ(k, k′) =

∫A

∫N

ψ(k′−1ank)ρ(a)1+s da dn dn .

Furthermore the operator πs(ψ) is of trace class and

trπs(ψ) =

∫K

qψ(k, k) dk =

∫A

∫N

∫K

ψ(kank−1) dk ρ(a)1+s dk dn da .

Proof. The formula for the integral kernel is obtained via a straightforward computationin [Lan85, page 48]. It is a continuous function. Hence the of trace class. �

By computing a few Jacobians, as in [Lan85, page 68], we obtain the following lemma.

Lemma 85. For φ ∈ Cc(G)∫N

φ(ana−1n−1) dn =1

|α(a)− 1|

∫N

φ(n) dn , (45)

and therefore ∫A\G

φ(x−1ax)) d.x =

ρ(a)

|D(a)|

∫N

∫K

φ(kank−1) dk dn , (46)

where

D(a) = ρ(a)− ρ(a)−1 (a ∈ A) . (47)

Theorem 86. Let Θπs be a function on G defined as follows.

Θπs(x) =

{2ρ(a)s+ρ(a)−s

|D(a)| if x is conjugate to a ∈ A ,

0 if x is not conjugate to any element of A .(48)

Then

trπs(ψ) =

∫G

Θπs(x)ψ(x) dx (ψ ∈ Cc(G)) . (49)

Proof. By combining Lemmas 84 and 85 we see that

trπs(ψ) =

∫A

|D(a)|∫

A\Gψ(x−1ax) d

.x ρ(a)s da . (50)

Changing the variable a to a−1 gives

trπs(ψ) =

∫A

|D(a−1)|∫

A\Gψ(x−1a−1x) d

.x ρ(a)−s da .

54 TOMASZ PRZEBINDA

However |D(a−1)| = |D(a)| and∫A\G

ψ(x−1a−1x) d.x =

∫A\G

ψ(x−1ax) d.x .

Therefore

trπs(ψ) =

∫A

|D(a)|∫

A\Gψ(x−1ax) d

.xρ(a)s + ρ(a)−s

2da .

On the other hand,∫G

Θπs(x)ψ(x) dx =1

2

∫A

|D(a)|2∫

A\GΘπs(x

−1ax)ψ(x−1ax) d.x da

=1

2

∫A

|D(a)|∫

A\Gψ(x−1ax) d

.xΘπs(a)|D(a)| da

and (48) follows. �

Notice that the trace is given via integration against a G-invariant locally integrablefunction. Let 0M = {1,−1} ⊆ G. This is the centralizer A in K. If δ is a character of thegroup 0M, we define V(δ, s) ⊆ V(s) to be the subspace of functions φ such that

φ(mx) = δ(m)ψ(x) (m ∈ 0M, x ∈ G) .

Denote by (πδ,s,V(δ, s)) the resulting representation of G. Clearly (πs,V(s)) is the directsum of the two subrepresentations (πδ,s,V(δ, s)).

Theorem 87. Let Θπδ,s be a function on G defined as follows.

Θπs(mx) =

{δ(m)ρ(a)s+ρ(a)−s

D(a)if x is conjugate to ma, where m ∈ 0M and a ∈ A ,

0 if x is not conjugate to any element of ±A .

Then

trπδ,s(ψ) =

∫G

Θπδ,s(x)ψ(x) dx (ψ ∈ Cc(G)) . (51)

Problem 17. Deduce Theorem 87 from Theorem 86.

12.4. Finite dimensional representations. Define

D(k(θ)) = eiθ − e−iθ (k(θ) ∈ K) . (52)

Lemma 88. Fix an integer n ≥ 1. For non-negative integers p,q with p + q = n − 1 letfp,q be a function of

x =

(a bc d

)such that fp,q(x) = cpdq. Then fp,q ∈ V(δ,−n), where δ(−1) = (−1)p+q. The functionsfp,q span a G-invariant finite dimensional subspace

U(δ,−n) =⊕

p+q=n−1

Cfp,q ⊆ V(δ,−n)


of dimension n. Denote the resulting representation by (σδ,−n,U(δ,−n)). Then

trσδ,−n(ma) = δ(m)ρ(a)n − ρ(a)−n

D(a)(m ∈ 0M,a ∈ A) . (53)

Also,

trσδ,−n(k(θ)) =einθ − e−inθ

D(k(θ))(k(θ) ∈ K) . (54)

Proof. Everything till (53) is straightforward. See [Lan85, page 151]. The formula (54)follows from (53). Indeed, let GC = SL2(C). Then G ⊆ GC is a subgroup and therepresentation (σδ,−n,U(δ,−n)) extends to a representation of GC so that σδ,−n : GC →GL(U(δ,−n)) is a polynomial map. Notice that

1√2

(1 −i1 i

)(cos θ sin θ− sin θ cosθ

)1√2

(1 1i −i

)=

(eiθ 00 e−iθ

)Thus k(θ) is conjugate to (

eiθ 00 e−iθ

)within GC. Hence Lemma 88 shows that

trσδ,−n(k(θ)) =einθ − e−inθ

D(k(θ))(k(θ) ∈ K) . (55)

�

Problem 18. Prove that for n = 2 the representation (σδ,−n,U(δ,−n)) is irreducible.(We’ll see soon that they are all irreducible.)

In this case this representation is isomorphic to the representation of G on C2 = M1,2(C)via right multiplication. Since no non-zero line in C2 is preserved by G, the claim follows.

Problem 19. Prove that for n ≥ 2 the representation (σδ,−n,U(δ,−n)) are not unita-rizable. (We know from Theorem 72 that this is the case. Try to find an independentargument.)

Each fp,q is an eigenfunction for the action of the subgroup A and the eigenvalues maygo to infinity. Hence, no matter what the inner product, the diagonal matrix coefficientdefined by fp,q is not bounded. This is impossible for a unitary representation

12.5. Smooth vectors and analytic vectors. Let (π,V) be a representation of G ona Banach space V. Denote by V∞ ⊆ V the subspace of all vectors v such that the map

G 3 x→ π(x)v ∈ V (56)

is smooth (infinitely many times differentiable). This subspace is dense because, as iseasy to check, π(C∞c (G))V ⊆ V∞. For X ∈ g = sl2(R), the Lie algebra of G, define

dπ(X)v =d

dtπ(exp(tX))v|t=0 (v ∈ V∞) . (57)

56 TOMASZ PRZEBINDA

One checks without difficulties that the map

dπ : g→ End(V∞) (58)

is a Lie algebra homomorphism. By linearity we extend it to the complexification gC =g + ig = sl2(C)

dπ : gC → End(V∞) (59)

and obtain a representation (dπ,V∞) of the Lie algebra gC.One drawback of this representation is that the closure of a g-invariant subspace U ⊆

V∞ in V does not need to be G-invariant. Indeed, consider the right regular representation(R,L2(G)) of G. Let U ⊆ G, U 6= G, be a non-empty open set. Then the space U =C∞c (U) is closed under the action of dR(g), but the closure L2(U) is not R(G) invariant.For this reason one is lead to study the space Van ⊆ V of all vectors v such that the map.

G 3 x→ π(x)v ∈ V (60)

is analytic.

Theorem 89. Let X ⊆ Van be a g-invariant vector subspace. Then the closure of X in Vis G-invarant

Proof. We follow [Var89, Theorem 2, page 108]. It is enough to prove that for any v ∈ Uand for any x ∈ G, π(x)u belonges to the closure of U. Suppose not. Then by Haan-Banach Theorem there is λ in the dual of V such that λ is equal to zero on the closure ofU, but for some x0 ∈ G, λ(π(x0))v 6= 0.

By assumption the function

G 3 x→ λ(π(x)v) ∈ C

is analytic. By the choice of λ, its Taylor series at x = 1 is zero. Indeed, for anyX1, X2, ..., Xn ∈ g

dπ(X1)dπ(X2)...dπ(Xn)v ∈ U

Hence

λ(dπ(X1)dπ(X2)...dπ(Xn)v) = 0

Also, λ(v) = 0. Thus the function is zero because G is connected. This is a contradiction.�

12.6. The derivative of the right regular representation. Every element g ∈ GL+(R)has a unique decomposition as

g =

(a bc d

)=

(u 00 u

)(y x0 1

)(cos θ sin θ− sin θ cos θ

), (61)

where u > 0, y > 0, x ∈ R and θ ∈ [−π, π). We extend any function defined on SL2(R)to a function on GL+(R) by making it independent of the variable u. Also, we extend the


right regular representation of SL2(R) to act on such defined functions on GL+(R). Thena straightforward computation, see [Lan85, 113-116] verifies the following formulas,

dR

(0 10 0

)= y cos 2θ ∂x + y sin 2θ ∂y + sin2 θ ∂θ , (62)

dR

(0 1−1 0

)= ∂θ ,

dR

(0 11 0

)= 2y cos 2θ ∂x + 2y sin 2θ ∂y − cos 2θ ∂θ ,

dR

(0 01 0

)= y cos 2θ ∂x + y sin 2θ ∂y − cos2 θ ∂θ ,

dR

(1 00 −1

)= −2y sin 2θ ∂x + 2y cos 2θ ∂y + sin 2θ ∂θ .

Problem 20. Verify the first two formulas in (62).

12.7. The universal enveloping algebra. Here g = sl2(R) and G = SL2(R). Theuniversal enveloping algebra U(g) is the complex tensor algebra of g divided by the idealgenerated by element AB−BA− [A,B], A,B ∈ g. The Poincare-Birkhoff-Witt Theoremsays that if A, B, C form a basis of the vector space gC, then the elements

AaBbCc (0 ≤ a, b, c ∈ Z)

form a basis of the vector space U(g). (Here A0 = B0 = C0 = 1.) Furthermore, for anyA,B,C ∈ gC satisfying the commutation relations

[A,B] = 2B , [A,C] = −2C , [B,C] = A

The element, called the Casimir element,

C = A2 + 2(BC + CB) (63)

generates the the center of U(g) and does not depend on the choice of the A, B, C.

Problem 21. Check that g acts trivially on C.

Thus the center of U(g) is equal to C[C]. Also, any representation of the Lie algebrag extends to a representation of the algebra U(g). In terms of the coordinates used insection 12.6,

dR(C) = 4y2(∂2x + ∂2

y)− 4y∂x∂θ (64)

Proofs may be found in [Lan85, pages 191-198].

12.8. K-multiplicity 1 representations. Here G = SL2(R) and we consider only rep-resentations (π,V) of G on Banach spaces V such that for each integer n the isotypiccomponent Vn ⊆ V has dimension at most 1.

Lemma 90. For any n ∈ Z, the space S∞n,n = Sn,n ∩ C∞(G) is dense in Sn,n

58 TOMASZ PRZEBINDA

Proof. See [Lan85, page 101] . �

Theorem 91. Let (π,V) be a representation G on a Banach space V such that for eachinteger n the isotypic component Vn ⊆ V has dimension at most 1. Then

VK =⊕n∈Z

Vn ⊆ Van .

In particular VK is a (g,K) module.

Proof. Since, by assumption, dimVn = 1, we see from Lemmas 90, 78 and 79 that

Vn = π(S∞n,n)Vn .

In particular, Vn ⊆ V∞. It’ll suffice to show that for a fixed n and a non-zero vectorv ∈ Vn the function

fv(g) = π(g)v (g ∈ G)

is analytic. Notice that

fv(gk(θ)) = einθfv(g) .

Hence (64) shows that

dR(C)fv(g) = (4y2(∂2x + ∂2

y)− 4y∂xin)fv(g) .

On the other hand dπ(C) commutes with π(K), hence preserves Vn. Since, by assumption,dimVn = 1, dπ(C) acts on Vn via multiplication by a scalar, call it cn. Thus

dπ(C)v = cnv .

Since

R(h)fv(g) = fπ(h)v(g) ,

this implies

dR(C)fvv = cnfv .

Since the characteristic variety of the system of differential equations

(4y2(∂2x + ∂2

y)− 4y∂xin)fv = cnfv , ∂θfv = in

is zero the function fv is analytic. �

Theorem 92. Let (π,V) be a representation G on a Banach space V such that for eachinteger n the isotypic component Vn ⊆ V has dimension at most 1. Then the map

V ⊇ U→ UK ⊆ VK

is a bijection between closed G-invariant subspaces of V and (g,K)-submodules of VK. Theinverse is given by

VK ⊇ X→ Cl(X) ⊆ V .

where Cl(X) denotes the closure of X in V. In particular (π,V) is irreducible if and onlyif the (g,K)-module VK is irreducible.

Proof. This is clear from Theorems 89 and 91. �


Lemma 93. For any irreducible (g,K)-module X and any integer n such that Xn 6= 0, anynon-zero vector v ∈ Xn is cyclic for the action of U(g) on X.

Proof. Since U(g)v is a submodule of X, the claim is obvious. �

Let X =⊕

n Xn ba a (g,K)-module with dimXn ≤ 1. A hermitian form (·, ·) on X iscalled g-invariant if

(Xu, v) = −(u,Xv) (u, v ∈ X, X ∈ g) . (65)

A (g,K) module is called unitarizable if it admits an invariant positive definite hemitianform.

Theorem 94. Let X =⊕

n Xn ba an irreducible (g,K)-module with dimXn ≤ 1. Then anytwo positive definite g-invariant hermitian products on X are positive multiples of eachother (assuming they exist).

Proof. Denote the two g-invariant positive hermitian products on X by (·, ·) and 〈·, ·〉.Then the spaces Vn are mutually orthogonal with respect to (·, ·) and 〈·, ·〉. Let Pn :V → Vn denote the orthogonal projection. Fix m ∈ Z such that Vm 6= 0 and a vector0 6= vm ∈ Vm. Lemma 93 implies that for any n there is a ∈ U(g) such that Pnavm 6= 0.Then

(avm, avm) = (vm, a∗avm) = (vm, Pma

∗avm) = c〈vm, Pma∗avm〉= c〈avm, avm〉 ,

where

c =(vm, vm)

〈vm, vm〉.

�

Theorem 95. Two irreducible unitary representations of G, of K-multiplicity at most 1,are unitarily isomorphic if and only if their (g,K)modules are isomorphic.

Proof. Let (π,V), (σ,U) be the two representations. If they are isomorphic, then clearlyso are the (g,K)-modules. Conversely, suppose

L : VK → UK

is a g-intertwinig map. Then (·, ·)V and the pull-back of (·, ·)U via L are positive definiteg-invariant hermitian products on VK. Theorem 94 shows that there is a constant c > 0such that

(v, v′)V = c(Lv, Lv′)U (v, v′ ∈ V) .

Hence

T =√cL : VK → UK

is a g-intertwining isometry. We need to check that T is also G-intertwining.

60 TOMASZ PRZEBINDA

Since K-finite vectors are analytic, we have

π(exp(X))v =∞∑n=0

1

n!(dπ(X))n v (v ∈ VK) .

For X ∈ g in some small neighborhood of zero. Hence, in this neighborhood,

Tπ(exp(X))v =∞∑n=0

1

n!(dσ(X))n Tv = σ(exp(X))Tv .

Thus there is an open neighborhood U of 1 ∈ G such that

Tπ(g)v = σ(g)Tv (g ∈ U) .

Since G is connected U generates G, so the proof is complete. �

12.9. The character of a (g,K)-module.

Theorem 96. Let (π,V) be a representation of G on a Hilbert space V, with dimVn ≤ 1for all n ∈ Z. Then for any φ ∈ C∞c (G), the operator π(φ) is of trace class and the map

C∞c (G) 3 φ→ trπ(φ) ∈ C

is a distribution on G.

Proof. Recall that∫G

φ(g)π(g) dg =

∫G

φ(g)π(gk−1k) dg =

∫G

R(k)φ(g)π(g) dg π(k) .

Hence by taking derivatives at k = 1,

0 =

∫G

dR(J)φ(g)π(g) dg +

∫G

φ(g)π(g) dg dπ(J) .

Let vn ∈ Vn be a unit vector. Then

0 =

∫G

dR(J)φ(g)π(g) dg vn +

∫G

φ(g)π(g) dg in vn .

By iterating we see that for m = 0, 1, 2, ...

(−in)mπ(φ)vn = π(dR(J)mφ) .

Thus

|(π(φ)vn, vn)| ≤ (1 + |n|)−m ‖ π(dR(J)mφ) ‖ (n ∈ Z) .

Since for m ≥ 2, ∑n∈Z

(1 + |n|)−m <∞ ,

the claim follows. �


Theorem 97. Let (π,V), (σ,U) be representations of G on Hilbert spaces V and U, withdimVn ≤ 1 and dim Un ≤ 1 for all n ∈ Z. Assume that the (g,K)-modules VK and UK

are isomorphic. Then,

trπ(φ) = tr σ(φ) (φ ∈ C∞c (G)) .

Proof. Given a linear map A : U→ V, recall the conjugate map T ∗ : V→ U,

(Au, v)V = (u,A∗v)U .

LetT : VK → UK

be a (g,K)-intertwining isomorphism. Pick unit vectors vn ∈ Vn. Notice that Tvn, n ∈ Z,form a basis of U and (T ∗)−1vn form a dual basis. Then

tr π(φ) =∑n∈Z

(π(φ)vn, vn)V and trσ(φ) =∑n∈Z

(σ(φ)Tvn, (T−1)∗vn)U .

By definition(σ(φ)Tvn, (T

−1)∗vn)U = (T−1σ(φ)Tvn, vn)U

Since vn is an analytic vector,

T−1σ(exp(X))Tvn = π(exp(X))

in a neighborhood of 0 ∈ g. Hence

(π(g)vn, vn)V = (σ(g)Tvn, (T−1)∗vn)U

in a neighborhood of 1 ∈ G. Since both functions are analytic, they are equal everywhere.Thus

(π(φ)vn, vn)V = (σ(φ)Tvn, (T−1)∗vn)U .

�

In general one says that two representations (π,V) and (σ,U) are infinitesimally equiv-alent if the (g,K)-modules VK and UK are isomorphic.

Given a representation of G (with K-multiplicities at most 1) we define the characterof π as

Θπ(φ) = tr π(φ) (φ ∈ C∞c (G) .

This is a distribution on G which, as shown in Theorem 97, does not depend on theinfinitesimal equivalence class of (π,V). Therefore we shall also write Θπ = ΘVK

.

12.10. The unitary dual.

Lemma 98. Let

J =

(0 1−1 0

), E+ =

(1 ii −1

), E− =

(1 −i−i −1

).

Then[E+, E−] = −4iJ , [J,E+] = 2iE+ , [J,E−] = −2iE− .

Proof. This is straightforward. See [Lan85, page 102]. �

62 TOMASZ PRZEBINDA

Corollary 99. For any (g,K) module X, with the K-isotypic components Xn,

JXn ⊆ Xn , E+Xn ⊆ Xn+2 , E−Xn ⊆ Xn−2 .

Proof. This is clear from Corollary 98 �

Recall the principal series representation (πs,V(s)).

Lemma 100. In terms of (61), for n ∈ Z define

vn

((u 00 u

)(y x0 1

)(cos θ sin θ− sin θ cos θ

))= y1+seinθ .

Then vn ∈ V(s) and

dπs(J)vn = invn ,

dπs(E−)vn = (s+ 1− n)vn−2 ,

dπs(E+)vn = (s+ 1 + n)vn+2 .

Proof. We see from (62) that

dR(J)vn = invn ,

dR(E−)vn = (s+ 1− n)vn−2 ,

dR(E+)vn = (s+ 1 + n)vn+2 .

But the right regular action coincides with πs, hence the formulas follow. �

By combining Lemma 100 with Theorem 91 we deduce the following Corollary.

Corollary 101. The (g,K) module of the principal series (πs,V(s)) is equal to

V(s)K =⊕n∈Z

Cvn .

(Here V(s) = L2(K), by restriction to K, independently of s and the restriction of vn toK is equal χn.)

Lemma 100 implies the following Proposition, see [Lan85, pages 119-121].

Proposition 102. If s is not an integer then

V(s)+K =

⊕n∈2Z

Cvn znd V(s)−K =⊕

n∈2Z+1

Cvn

are irreducible submodules of V(s)K and V(s)K is the direct sum of them.If s = 0, then V(0)K is the dirct sum of three irreducible submodules

V(0)K =⊕n∈2Z

Cvn ⊕⊕

1≤n∈2Z+1

Cvn ⊕⊕

−1≥n∈2Z+1

Cvn .

If m ≥ 2 is an integer and s = m−1, then V(m−1)K contains three irreducible submodules

Xm =⊕

m≤n, n−m∈2Z

Cvn , X−m =⊕

−m≥n, n−m∈2Z

Cvn ,⊕

n−m∈2Z+1

Cvn .


The quotient module, V(m − 1)K divided by the three submodules is irreducible, finitedimensional of dimension m− 1. It has a basis represented by the elements

v−m+2 , v−m+4 , ... , vm−2 .

If m ≥ 2 is an integer and s = −m + 1, then V(m− 1)K contains the finite dimensionalsubmodule

Cv−m+2 ⊕ Cv−m+4 ⊕ ...⊕ Cvm−2 .

The quotient module, V(m − 1)K divided by this module is isomorphic to the direct sumof modules Xm and X−m plus the sum of all K-types of parity opposite to m.

The (g,K)-modules V(m− 1)K and V(−m+ 1)K are dual to each other.

Thus we have the highest weight modules, lowest weight modules, finite dimensionalmodules and modules with unbounded K-types on both side.

Problem 22. Write down a proof of Theorem 102. Imagine studing the structure ofthe principal series without the notion of a (g,K)-module and without Harish-Chandra’sreduction.This is written in Lang’s book.

Proposition 103. The Casimir element C, (63), acts on the principal series representation(πs,V(s)) via multiplication by s2 − 1:

dπs(C) = (s2 − 1)I .

Proof. As checked in [Lan85, page 195],

C = −1− (J − i)2 + E+E− .

Hence a straightforward computation using Lemma 100 implies the formula. �

In particular, since the discrete series modules Xn+1, X−n−1 are contained in a principalseries V(n)K we see that the Casimir element acts on them via multiplication by n2 − 1.Similarly the Casimir element acts on an irreducible finite dimensional representation ofdimension n via multiplication by n2 − 1.

The commutation relations Lemma 98 and the formulas Lemma 100 with some workimply the following theorem, due to Bargmann, [Bar47]. See [Lan85, page 123].

Theorem 104. Here is a complete list of the irreducible (g,K) modules, up to equivalence.

(1) Lowest weight module Xm with lowest weight m ≥ 1 and the highest weight moduleXm with highest weight m ≤ −1

(2) Principal series V(iτ)+K and V(iτ)−K, τ ∈ R \ {0};

(3) Principal series V(0)+K;

(4) Complementary series V(s)+K, −1 < s < 1;

(5) Trivial module.

Problem 23. Based on Lang, write down a proof of this theorem. Also, deduce fromit that the only finite dimensional irreducible unitary representation of G is the trivialrepresentation.

64 TOMASZ PRZEBINDA

A similar result was obtained by Dan Barbasch, [Bar89] for the complex classical groups.The closures of these modules in the corresponding principal series representation are

representations of G on Hilbert spaces. They are unitary representations except thehighest and lowest weight representations. In these cases the inner product inheritedfrom the principal series is not g invariant. Therefore there is a problem of constructingthe unitary representations of G whose (g,K)-modules are the highest and lowest weightmodules. This is explained in the theorem below. See [Lan85, page 181].

Theorem 105. Define

α

(a bc d

)=

1

2(a+ d− ic+ ib) , β

(a bc d

)=

1

2(c+ b− ia+ id) .

Let m ≥ 2. Then the closure V(m) in L2(G) of the space

∞⊕r=0

Cα−m−rβr

is invariant under the left action

L(g)φ(x) = φ(g−1x) .

The resulting representation (L,V(m)) of G is irreducible and unitary. Its (g,K)-module

V(m)K is isomorphis to the lowest weight module Xm with the lowest weight m. In order to

realize the lowest weight representations we take the complex conjugate of V(m).

We shall skip the construction of the unitary representations whose (g,K)-modules areX1 and X−1. They are not square integrable. One may find them in [Kna86, page 36].

12.11. The character of the sum of discrete series. For an integer m ≥ 2 let ΘX(m)

denote the the character of the (g,K)-module X(m), and ΘX(−m) denote the the characterof the (g,K)-module X(−m). Let

ht =

(et 00 e−t

)t ∈ R) .

Proposition 106. The character of the sum of the two discrete series (g,K)-modules X(m)

and X(−m) is represented by the G-conjugation invariant function ΘX(m)⊕X(−m) given by

ΘX(m)⊕X(−m)(k(θ)) = −ei(m−1)θ − e−i(m−1)θ

eiθ − e−iθ(θ ∈ R) ,

ΘX(m)+X(−m)(zht) = zme−t|m−1|

|et − e−t|(z = ±1, t ∈ R) .

Proof. From Proposition 102 we know the structure of the principal series V(m−1)K, fromTheorem 86, the character of it, from Lemma 88 the character of the finite dimensional


representation which is in the principal series. Now Theorem 97 justifies the formula

character of principal series

= character of the sum of discrete series + character of finite dimensional module .

This completes the proof �

For convenience define The following G-invariant functions on G

ΘX(m)(k(θ)) = − ei(m−1)θ


ΘX(m)(zht) = zme−|t||m−1|

|et − e−t|(z = ±1, t ∈ R) .

and

ΘX(−m)(k(θ)) =ei(m−1)θ


ΘX(−m)(zht) = zme−|t||m−1|

|et − e−t|(z = ±1, t ∈ R) .

These are the characters of the individual discrete series representations, but we don’tneed to know it to explain Harish-Chandra’s Plancharel formula in the SL2(R) case.

12.12. Harish-Chandra’s Plancherel formula for L2(G). It is not difficult to checkthat K and H = A ∪ (−A) are the only Cartan subgroups of G up to conjugacy. LetA′ = {a ∈ A; a 6= 1} and let K′ = {t ∈ K; t 6= ±1}. We begin by recalling Harish-Chandra’s orbital integrals of a function φ ∈ C∞c (G), using the notation of [Lan85],

HAφ (zht) = |D(ht)|

∫A\G

φ(x−1zhtx) d.x (z = ±1, t ∈ R, t 6= 0) , (66)

HKφ (k) = D(k)

∫K\G

φ(x−1kx) d.x (k ∈ K′) ,

The Haar measure on G may be expressed in terms of these integrals by∫G

φ(x) dx =

∫K

HKφ (k)D(k) dk +

1

2

∫A′HAφ (a)|D(a)| da+

1

2

∫A′HAφ (−a)|D(a)| da . (67)

Theorem 107. The function HAφ extends to a smooth function on H. The function HK

φ issmooth on H′ and its derivatives have one sided limits on the boundary. In these terms,

HKφ (k(0+))−HK

φ (k(0−)) =i

2HAφ (1) , (68)

HKφ (k(π+))−HK

φ (k(π−)) =i

2HAφ (−1)) ,

Furthermore,

∂θHKφ (k(θ))|θ=0 = −iφ(1) . (69)

66 TOMASZ PRZEBINDA

Proof. This is a problem concerning integrals on a 3 dimensional manifold, G = SL2(R).Notice that we may replace φ by a K conjugation invarian function

∫Kφ(kxk−1) dk. This

leads to analysis on the two dimensional manifold G/K. The computations are done in[Lan85, page 164 -167]. �

Problem 24. Take a look at Lemma on page 164 in Lang’s book to understand the kindof elementary calculus we are using here. Harish-Chandra proved an analogous statementfor an arbitrary semisimple group. Imagine the difficulties.

Recall the character χn(k(θ)) = einθ.

Theorem 108. For any integer m ≥ 2 and φ ∈ C∞c (G)∫G

ΘX(m)(x)φ(x) dx =

∫K

HKφ (k)χm−1(k) dk

+1

2

∫RHAφ (ht)e

−|t||m−1| dt+1

2

∫R(−1)mHA

φ (−ht)e−|t||m−1| dt

and∫G

ΘX(−m)(x)φ(x) dx = −∫

K

FKφ (t)χ−m+1(k) dk

+1

2

∫RHAφ (ht)e

−|t||m−1| dt+1

2

∫R(−1)mHA

φ (−ht)e−|t||m−1| dt .

Proof. This follows from (67) and the formulas for ΘX(±m) in previous section. �

Theorem 108 gives formulas for the Fourier coefficients of HKφ :

HKφ (n) =

∫K

HKφ (k)χn(k) dk (n 6= 0) .

Indeed, for n ≥ 1,

HKφ (n) = ΘX(n+1)(φ)− 1

2

∫RHAφ (ht)e

−|t|n dt− 1

2

∫R(−1)mHA

φ (−ht)e−|t|n dt

and

HKφ (−n) = ΘX(−n−1)(φ)− 1

2

∫RHAφ (ht)e

−|t||n| dt− 1

2

∫R(−1)mHA

φ (−ht)e−|t||n| dt

In particular,

HKφ (n)− HK

φ (−n) = ΘX(n+1)+X(−n−1)(φ)

−∫RHAφ (ht)e

−|t|n dt−∫R(−1)mHA

φ (−ht)e−|t|n dt . (70)

On the other hand, for k 6= ±1

HKφ (k) =

∑n∈Z

HKφ (n)χ−n(k) =

∑n∈Z

HKφ (n)χn(k−1) =

∑n∈Z

HKφ (−n)χ−n(k−1)


and therefore

HKφ (k)−HK

φ (k−1) =∑

06=n∈Z

(HKφ (n)−HK

φ (−n))χ−n(k) .

Thus,

HKφ (k(θ))−HK

φ (k(−θ)) =∞∑n=1

(HKφ (n)−HK

φ (−n))(−i) sin(nθ) . (71)

Continuing this way (and correcting the constants, if necessary) one obtains the followinglemma, see [Lan85, page 174],

Lemma 109.

π

i(HK

φ (k(θ))−HKφ (k(−θ))) = −

∞∑n=1

ΘX(n+1)+X(−n−1)(φ) sin(nθ)

+∞∑n=1

∫R

1

2

(HAφ (ht) + (−1)n+1HA

φ (−ht))

sin(nθ)e−|t|n dt . (72)

Let

φ+(x) =φ(x) + φ(−x)

2and φ−(x) =

φ(x)− φ(−x)

2.

Then a straightforward argument shows that

∞∑n=1

∫R

1

2

(HAφ (ht) + (−1)n+1HA

φ (−ht))

sin(nθ)e−|t|n dt

=

∫RHAφ+(ht)

sin(θ) cosh(t)

cosh(2t)− cos(2θ)dt+

∫RHAφ+(ht)

sin(2θ)

cosh(2t)− cos(2θ)dt

For ψ ∈ C∞c (G) we have the Fourier transform

HAψ (λ) =

∫RHAψ (ht) e

itλ dt = Θπiλ(φ) , (73)

where the second equality follows from (50). One computes the Fourier transforms of thefunctions

sin(θ) cosh(t)

cosh(2t)− cos(2θ)and

sin(2θ)

cosh(2t)− cos(2θ)

to deduce from Lemma (109) and (73) the following theorem. See [Lan85, page 173]

Theorem 110.

π

i(HK

φ (k(θ))−HKφ (k(−θ))) = −

∞∑n=1

ΘX(n+1)+X(−n−1)(φ) sin(nθ)

+1

2

∫ ∞0

Θπ+,iλ(φ)cosh((π

2− θ)λ)

cosh(πλ2

)dλ+

1

2

∫ ∞0

Θπ−,iλ(φ)sinh((π

2− θ)λ)

sinh(πλ2

)dλ

68 TOMASZ PRZEBINDA

Now we take the derivative with respect to θ of both sides, go to limit with θ → 0 andapply (69) to deduce the Harish-Chandra’s Plancherel formula.

Theorem 111. For any φ ∈ C∞c (G),

2πφ(1) =∞∑n=1

nΘX(n+1)+X(−n−1)(φ)

+1

2

∫ ∞0

Θπ+,iλ(φ)λ tanh(πλ

2) dλ+

1

2

∫ ∞0

Θπ−,iλ(φ)λ coth(πλ

2) dλ .

This theorem was published by Harish-Chandra in 1952, [HC52]. Notice that knowingthe composition series of the principal series was crucial for this proof.

Since Θπ(φ) = tr π(φ) the formula of Theorem 111 may be rewritten as

φ(1) =1

2π

∞∑n=1

n (tr(πn+1(φ)) + tr(π−n−1(φ)))

+1

4π

∫ ∞0

trπ+,iλ(φ)λ tanh(πλ

2) dλ+

1

4π

∫ ∞0

trπ−,iλ(φ)λ coth(πλ

2) dλ , (74)

where πm is the discrete series representation of G with the (g,K)-module X(m). Let usfix x ∈ G and replace φ by R(x)φ. Then

R(x)φ(1) = φ(x)

and

π(R(x)φ) =

∫G

φ(gx)π(g) dg =

∫G

φ(g)π(gx−1) dg =

∫G

φ(g)π(g) dgπ(x−1) = π(φ)π(x−1) .

Hence,

φ(x) =1

2π

∞∑n=1

n(tr(πn+1(φ)π(x−1)) + tr(π−n−1(φ)π(x−1))

)+

1

4π

∫ ∞0

tr(π+,iλ(φ)π(x−1))λ tanh(πλ

2) dλ+

1

4π

∫ ∞0

tr(π−,iλ(φ)π(x−1))λ coth(πλ

2) dλ .

Therefore one defines the Fourier transform of φ to be an operator valued function

F(φ)(π) = π(φ) =

∫G

φ(g)π(g) dg ∈ End(Vπ) (φ ∈ C∞c (G)) , (75)

where π is one of the irreducible unitary representations of G which occur in Theorem(111), realized on a Hilbert space Vπ. Furthermore we have the Hilbert-Schmidt innerproduct and norm on the subspace End(Vπ)HS ⊆ End(Vπ),

(S, T ) = tr(ST ∗) , ‖ T ‖22= (T, T ) (S, T ∈ End(Vπ)) .

With this notation, we have the following Fourier Inversion Formula



φ(x) =1

2π

∞∑n=1

n(tr(F(φ)(πn+1)π(x−1)) + tr(F(φ)(π−n−1)π(x−1))

)+

1

4π

∫ ∞0

tr(F(φ)(π+,iλ)π(x−1))λ tanh(πλ

2) dλ+

1

4π

∫ ∞0

tr(F(φ)(π−,iλ)π(x−1))λ coth(πλ

2) dλ .

Recall also that φ∗(x) = φ(x−1) so that

‖ φ ‖22=

∫G

|φ(x)|2 dx =

∫G

φ(x)φ∗(x−1) dx =

∫G

φ(x)φ∗(x−11) dx = φ ∗ φ∗(1) .

Also, for a unitary representation π,

π(φ)∗ =

(∫G

π(g)φ(g) dg

)∗=

∫G

π(g)∗φ(g) dg =

∫G

π(g−1)φ(g) dg =

∫G

π(g)φ(g−1) dg

= π(φ∗) .

Hence (74) implies the following theorem


‖ φ ‖22=

1

2π

∞∑n=1

n(‖ F(φ)(πn+1) ‖2

2 + ‖ F(φ)(π−n−1) ‖22

)+

1

4π

∫ ∞0

‖ F(φ)(π+,iλ) ‖22 λ tanh(

πλ

2) dλ+

1

4π

∫ ∞0

‖ F(φ)(π−,iλ) ‖22 λ coth(

πλ

2) dλ .

Also, for ψ ∈ C∞c (G),

(φ, ψ) =1

2π

∞∑n=1

n ((F(φ)(πn+1),F(ψ)(πn+1)) + (F(φ)(π−n−1),F(ψ)(π−n−1)))

+1

4π

∫ ∞0

(F(φ)(π+,iλ),F(ψ)(π+,iλ))λ tanh(πλ

2) dλ

+1

4π

∫ ∞0

(F(φ)(π−,iλ),F(ψ)(π−,iλ))λ coth(πλ

2) dλ .

Recall, [Kir78, page 68] the notion of the direct integral of Hilbert spaces Hx, indexedby x in some set X equipped with a measure µ,∫

X

Hx dµ(x) .

This is a Hilbert space of functions f such that for each x ∈ X, f(x) ∈ Hx and

(f, h)∫X Hx dµ(x) =

∫X

(f(x), h(x))Hx dµ(x) .

70 TOMASZ PRZEBINDA

We take X to be the set Gtemp of the equivalence classes of the irreducible unitary repre-

sentations of G which occur in Theorem 111 and for each π ∈ Gtemp, Hπ = End(Vπ). Wemay identify

Gtemp = (Z \ {0}) ∪ ({±} × R+)

and define the measure

dµ(πn) = dµ(π−n) = dµ(n) =1

2π|n| (n ∈ Z \ {0}) ,

dµ(π+,λ) = dµ(+, λ) =1

4πλ tanh(

πλ

2) dλ (λ ∈ R+) ,

dµ(π−,λ) = dµ(−, λ) =1

4πλ coth(

πλ

2) dλ (λ ∈ R+) .

The subspace C∞c (G) ⊆ L2(G) inherits the inner product and Theorem 113 impliesthat the Fourier transform is an injective isometry of Hilbert spaces

F : C∞c (G)→∫

Gtemp

End(Vπ)HS dµ(π) . (76)

The group G×G acts on C∞c (G) via the left and right regular representation:

φ(x)→ φ(g−1xh) (g, h, x ∈ G) .

This groups also acts on the direct integral of Hilbert spaces, (76),

End(Vπ)HS 3 T → π(g)Tπ(h)−1 ∈ End(Vπ)HS (π ∈ Gtemp) (77)

and it is easy to see that the Fourier transform (76) intertwines these two actions. Clearlyboth actions are by unitary representations. With some more effort one deduces thefollowing theorem.

Theorem 114. The Fourier transform (76) extends to a G × G intertwining bijectiveisometry of Hilbert spaces

F : L2(G)→∫

Gtemp

End(Vπ)HS dµ(π) . (78)

Recall that the map

Vπ ⊗ Vπ 3 u⊗ v → Eu⊗v ∈ End(Vπ) ,

Eu⊗v(w) = (w, u)Vπv

extends to a bijective isometry. It is linear with respect to v and anti-linear with respectto u. Moreover,

π(g)Eu⊗vπ(h)−1(w) = (π(h)−1w, u)Vππ(g)v = (w, π(h)u)Vππ(g)v = Eπ(h)u⊗π(g)v .

If we introduce a new multiplication by complex numbers on the “left” copy of Vπ by

z · u = zu (z ∈ C)


then we still have z · π(h)u = π(h)z · u. Thus the new Vπ, call it Vπ still carries a unitaryrepresentation of G, call it π. The result is a representation (π,Vπ). Altogether we havea G×G intertwining bijective isometry which extends the map

Vπ ⊗ Vπ 3 u⊗ v → Eu⊗v ∈ End(Vπ) .

Theorem 115. The Fourier transform (76) extends to a G × G intertwining bijectiveisometry of Hilbert spaces

F : L2(G)→∫

Gtemp

Vπ ⊗ Vπ dµ(π) . (79)

Symbolically

F : (L⊗R,L2(G))→∫

Gtemp

(π ⊗ π,Vπ ⊗ Vπ) dµ(π) , (80)

or

L⊗R =

∫Gtemp

π ⊗ π dµ(π) . (81)

12.13. An irreducible unitary non-tempered representation of SL2(R).

12.13.1. An abstract (g,K)-module. We begin by constructing an irreducible unita-rizable (g,K)-modules which is infinite dimensional and does not occur on the list of(g,K)-modules which are present in the Harish-Chandra Plancherel formula, Theorem111.

Let

e+ =

(0 10 0

), e− =

(0 01 0

), h =

(1 00 −1

).

Then[e+, e−] = h , [h, e+] = 2e+ , [h, e−] = −2e−

andg = Re+ ⊕ Re− ⊕ Rh .

Let

~ = i(e− − e+) , n+ =1

2(h+ i(e+ + e−)) , n− =

1

2(h− i(e+ + e−)) .

Then[n+, n−] = ~ , [~, n+] = 2n+ , [~, n−] = −2n−

andgC = Cn+ ⊕ Cn− ⊕ C~ .

Moreover~ = −~ , n+ = n− .

As a vector space

X =∞⊕j=0

Cvj ,

72 TOMASZ PRZEBINDA

where v0, v1, ... is a basis of X and the action of gC is given by

vj = (n+)jv0 (j = 0, 1, 2, ...) ,

n−vj = −j2vj−1 (j = 1, 2, ...) ,

n−v0 = 0 ,

~vj = (1 + 2j)vj (j = 0, 1, 2, ...) .

The group K = SO2 acts by taking the exponential of the last formula

k(θ)vj = ei(1+2j)θvj (j = 0, 1, 2, ...) .

Indeed,

k(θ)vj = exp(θ((e+ − e−))vj

and

θ((e+ − e−)vj = iθ(i(e− − e+))vj = iθ(1 + 2j)vj .

Hence the formula follows. Thus X is a (g,K)-module and it is not difficult to check thatit is irreducible.

We need to define an invariant positive definite Hermitian product (·, ·) on X. Thus wemust have

((X + iY )u, v) = (u,−(X + iY )v) (X, Y ∈ g; u, v ∈ X) .

We do it as follows. We declare the vj to be orthogonal and set

(vj, vj) = (j!)2 (j = 0, 1, 2, ...) .

Then

(vj+1, vj+1) = (n+vj, vj+1) = (vj,−n−vj+1) = (vj, (j + 1)2vj) = (j + 1)2(vj, vj) .

This implies that the inner product is invariant indeed.

12.13.2. A realization of the abstract (g,K)-module. Next we realize the (g,K)-module explicitly.

Define te following operators on the space S(R2),

ω(h) = x1∂x1 + x2∂x2 + 1 ,

ω(e+) =i

2(x2

1 + x22) ,

ω(e−) =i

2(∂2x1

+ ∂2x2

) .

The resulting map ω : g → End(S(R2)) defines a representation of the Lie algebra gC.Let aj = xj + ∂xj , bj = xj − ∂xj , ∆ = ∂2

x1+ ∂2

x2and let r2 = x2

1 + x22 A straightforward


computation shows that

ω(~) =1

4

2∑j=1

(ajbj + bjaj) ,

ω(n+) =i

4er2

2 ∆e−r2

2 ,

ω(n−) =i

4e−

r2

2 ∆er2

2 .

Let v0 = 1√πe−

r2

2 and let (·, ·) denote the usual L2 inner product on S(R). Then

(v0, v0) = 1 .

We can realize the (g,K)-module constructed previously as

X =∞⊕j=0

Cvj ⊆ S(R2) .

where vj = ω(n+)jv0, j = 0, 1, 2, ....

12.13.3. The irreducible unitary representation of the group. Finally we describethe representation of the group G whose (g,K)-module is equal to X. The orthogonalgroup O2 acts on S(R2) in the usual fashion

ω(g)v(x) = v(xg) (v ∈ S(R2), x ∈ R2, g ∈ O2) .

Let S(R2)O2 ⊆ S(R2) denote the subspace of the invariants. Recall the Iwasawa decom-position G = KAN, (38). For v ∈ S(R2) set

ω(

(a 00 a−1

))v(x) = av(ax) ,

ω(

(1 n0 1

))v(x) = ein

r2

2 v(x) ,

ω(

(0 1−1 0

))v(x) =

1

2π

∫R

∫Rei(x1y1+x2y2)v(y1, y2) dy1 dy2 .

Since he space S(R2)O2 is dense in L2(R2)O2 and since the subgroups A, N together

with the element

(0 1−1 0

)generate G we conclude that (ω,L2(R2)O2) is an irreducible

unitary representation of G with the (g,K)-module X desribed above.

13. The Heisenberg group.

13.1. Structure of the group. Let W be a finite dimensional vector space over R witha non-degenerate symplectic form 〈 , 〉. This means that 〈 , 〉 is a bilinear form such that

〈w,w′〉 = −〈w′, w〉

74 TOMASZ PRZEBINDA

and if

〈w,w′〉 = 0

for all w′ ∈ W then w = 0. As is well known, the dimension of W is even and W has abasis

e1, e2, ..., en, f1, f2, ..., fn

such that

〈ej, fk〉 = δj,k , 〈ej, ek〉 = 〈fj, fk〉 = 0 .

By fixing such a basis we get linear isomorphism

R2n 3 (w1, ..., wn, wn+1, .., w2n)→ w =n∑j=1

wjej +n∑k=1

wn+kfk ∈ W . (82)

Moreover the formula

J(ej) = −fj , J(fk) = ek

defines a linear map J ∈ End(W) such that J2 = −I and the bilinear form

〈J ·, ·〉 (83)

is symmetric and positive definite. Any map with these properies is calle a positive definitecompatible complex structure on W.

Using the identification (82) we define a Lebesgue measure dw on W by

dw = dw1...dw2n

It is characterized by the property that the volume of the unit cube with respect to theform (83) is 1.

Furthermore, the two subspaces

X =n∑j=1

Rej , Y =n∑j=1

Rfj

are maximal isotropic and

W = X⊕ Y . (84)

By the Heisenberg group we understand the direct product H = H(W) = W×R with themultiplication given by

(w, r)(w′, r′) = (w + w′, r + r′ +1

2〈w,w′〉) ((w, r), (w′, r′) ∈ H(W)).

Then Z = {0} × R is the center of H.


13.2. Decomposition of the right regular representation. For f1, f2 ∈ S(H) wehave the convolution

f1 ∗ f2(w, r) =

∫H

f1(w′, r′)f2((w, r)(w′, r′)−1) dw′ dr′

=

∫H

f1(w′, r′)f2((w, r)(−w′,−r′)) dw′ dr′

=

∫H

f1(w′, r′)f2((w − w′, r − r′ − 1

2〈w,w′〉) dw′ dr′

Define the central Fourier transform

ZF : S(H)→ S(H)

by the formula

ZFf(w, χ) =

∫Rf(w, r)χ(r) dr , (85)

where χ is the unitary character of R identified with a real number χ by χ(r) = e2πiχr,r ∈ R. A straightforward computation shows that

ZF(f1 ∗ f2)(w, χ) =

∫W

ZFf1(w′, χ)ZFf2(w − w′, χ)χ(1

2〈w,w′〉) dw′ . (86)

Thus if we introduce the following twisted convolution in S(W)

ψ\φ(w) =

∫W

ψ(u)φ(w − u)χ(1

2〈u,w〉) du (w ∈ W) (87)

then the central Fourier transform followed by evaluation at χ reults in the twisted con-volution of the central Fourier transforms evaluated at χ. Also,

ZFf ∗(w, χ) = ZFf(w, χ) . (88)

Plancherel formula for R shows that

‖ f ‖22=‖ ZFf ‖2

2 . (89)

Define the Weyl transform

Kχ : S(W)→ S(X× X) , (90)

Kχ(φ)(x, x′) =

∫Y

φ(x− x′ + y)χ(1

2〈y, x+ x′〉) dy .

A straightforward computation shows that

Kχ(φ\ψ)(x, x′′) =

∫X

Kχ(φ)(x, x′)Kχ(ψ)(x′, x′′) dx′ (91)

and

‖ Kχ(φ) ‖22= |χ|−n ‖ φ ‖2

2 . (92)

76 TOMASZ PRZEBINDA

Problem 25. Verify (91) and (92).

To check (92) we compute,∫X

∫X

|Kχ(φ)(x, x′)|2 dx dx′ = 2−n∫X

∫X

|Kχ(φ)(u+ v

2

, u− v2

)|2 du dv

2−n∫X

∫X

|∫

Y

φ(v + y)χ(1

2〈y, u〉) dy|2 du dv = 2−n

∫X

∫X

|∫

Y

φ(v + y)eiπχ〈y,u〉 dy|2 du dv

=

∫X

∫X

|∫

Y

φ(v + y)e2πiχ〈y,u〉 dy|2 du dv =

∫X

∫X

|∫

Y

φ(v + y)e2πi〈y,χu〉 dy|2 du dv

= |χ|−n∫X

∫X

|∫

Y

φ(v+y)e2πi〈y,u〉 dy|2 du dv = |χ|−n∫X

∫X

|φ(v+u)|2 du dv = |χ|−n ‖ φ ‖22 ,

where in the second to last equation we use the fact that the Fourier transform on Rn isan isometry.

SetZFχ(f)(w) = ZF(f)(w, χ)

Then our computations show that∫R‖ Kχ(ZFχ(f)) ‖2

2 |χ|n dχ =‖ f ‖22 . (93)

Each element K ∈ S(X× X) defines an operator Op(K) ∈ Hom(S(X),S(X)) by

Op(K)v(x) =

∫X

K(x, x′)v(x′) dx′ . (94)

Recall the Hilbert-Schmidt norm for operators on L2(X),

‖ Op(K) ‖2H.S.=

∫X

∫X

|K(x, x′)|2 dx dx′

Then (93) may be rewritten as∫R‖ Op ◦ Kχ ◦ ZFχ(f)) ‖2

H.S. |χ|n dχ =‖ f ‖22 (f ∈ S(H)) . (95)

Setωχ = Op ◦ Kχ ◦ ZFχ .

Then, from what we just computed

ωχ : S(H(W))→ End(S(X))

is an injective ∗-algebra homomorphism. By using approximate identity sequences onS(H) we extend ωχ to a map

ωχ : H(W)→ End(S(X)) .

LetU(L2(X))


denote the group of the unitary operators on the Hilbert space L2(X).

Theorem 116. The map

ωχ : H(W)→ U(L2(X)) ∩ End(S(X))

is an injective group homomorphism. For each v ∈ L2(X), the map

H(W) 3 g → ωχ(g)v ∈ L2(X)

is continuous, so that (ωχ,L2(X)) is a unitary representation of the group. Explicitly, for

v ∈ L2(X),

ωχ(x0, r)v(x) = χ(r)v(x− x0) (x, x0 ∈ X, r ∈ R) , (96)

ωχ(y0, r)v(x) = χ(r)χ(〈y0, x〉)v(x) (y0 ∈ Y, r ∈ R) ,

Hence, the representation (ωχ,L2(X)) of H(W) is irreducible. Moreover

‖ f ‖22=

∫R‖ ωχ(f)) ‖2

H.S. |χ|n dχ . (97)

Hence, |χ|n dχ is the Plancherel measure and

(L,L2(H(W))) =

∫R\{0}

(ωχ,L2(X))|χ|n dχ . (98)

Proof. We’ll check that the representation (ωχ,L2(X)) given by the formula (96) is irre-

ducible. Suppose T : L2(X) → L2(X) is a bounded operator that commutes with ωχ(g)for all g ∈ H(W). In particular

T : χ(〈y0, x〉)v(x)→ χ(〈y0, x〉)Tv(x) (y0 ∈ Y) .

Hence, for any f ∈ S(Y),

T :

∫Y

f(y0)χ(〈y0, x〉) dy0 v(x)→∫Y

f(y0)χ(〈y0, x〉) dy0 Tv(x) (y0 ∈ Y) .

Set ∫Y

f(y0)χ(〈y0, x〉) dy = f(x) .

Then f is a Fourier transform of f . Thus we checked that T commutes with the mul-tiplication by f . Since the Fourier transform is bijective on the space of the Schwartzfunctions, we see that T commutes with the multiplication by any Schwartz function,

v(x)→ f(x)v(x) (x ∈ X, f ∈ S(X), v ∈ L2(X)) .

Let vR be the indicator function of the ball BR of radius R centered at the origin. Thenfor any v ∈ C∞c (X) supported in that ball,

T (v) = T (vRv) = vT (vR) = T (vR)v .

Hence T (vR) is a bounded measurable function and

T (v) = T (vR)v (v ∈ C∞c (BR)) .

78 TOMASZ PRZEBINDA

This implies thatT (vR′)|BR = T (vR) (R < R′) .

Hence there is a function F such that

T (v)(x) = F (x)v(x) (v ∈ C∞c (X)) .

But we also know that T commutes with the translations. Hence F is a constant function.Therefore the representation is irreducible. �

13.3. Structure of the Lie algebra. The Lie algebra h(W) of the Heisenberg group isthe direct sum h(W) = W⊕R with the obvious structure of the vector space over R andthe Lie bracket given by

[(w, r), (w′, r′)] = (0, 〈w,w′〉) ((w, r), (w′, r′) ∈ h(W)).

Then, under the identification of manifolds, h(W) = H(W), the commutator in the groupand the Lie bracket in the Lie algebra are related by the formula

(w, r)(w′, r′)(w, r−1)(w′, r′)−1 = [(w, r), (w′, r′)] .

The commutators of higher order are trivial. Hence H(W) is a step two nilpotent Liegroup. Similarly, h(W) is a step two nilpotent Lie algebra.

13.4. A generic representation of the Lie algebra. Let us fix a nontrivial characterχ as in Theorem 116. Then by taking the derivatives we obtain the following action ofthe Lie algebra

ωχ(x0, 0)v(x) = −∂x0v(x) (x, x0 ∈ X, v ∈ S(X)) , (99)

ωχ(y0, 0)v(x) = 2πiχ〈y0, x〉v(x) (x ∈ X, y0 ∈ Y, v ∈ S(X)) ,

ωχ(0, r)v(x) = 2πiχrv(x) (r ∈ R, v ∈ S(X)) .

where ∂x0 is the directional derivative in the direction of x0

∂x0v(x) = limt→0

(v(x+ tx0)− v(x))t−1 .

Let us identify X with Rn, by

X 3n∑j=1

njej = (x1, x2, ..., xn) ∈ Rn .

Also, lets choose χ so that 2πχ = 1. Set ω = ωχ. Then

ωχ(ej) = −∂j , ωχ(fj) = ixj (1 ≤ j, k ≤ n) . (100)

This is the “easiest on the eyes” version of the Heisenberg representation of the HeisenbergLie algebra with its Canonical Commutation Relations

[∂j, xk] = δj,k .

14. An example of a wild group.

Here we follow [Kir78, $19]. This is the last section of this book. It contains some“removable singularities”. Please check comparing with the text below.


14.1. The Mautner group. Fix an irrational number α ∈ R. Let G be the group ofmatrices eit 0 z

0 eiαt w0 0 1

,

where t ∈ R, z, w ∈ C. Let us identity the above matrix with the string of numbers(t, z, w). Then t

(t, z, w)(τ, ξ, η) = (t+ τ, z + eitξ, w + eiαtη) ,

(t, z, w)(0, 0, 0) = (t, z, w) ,

(τ, ξ, η)−1 = (−τ,−e−iτξ,−e−iατη) .

If we identify R = {(t, 0, 0); t ∈ R} and C2 = {(0, z, w); z, w ∈ C}, then we see that G isthe semidirect product of R and C2. Thus G is a solvable group. It is easy to check thatthe formula ∫

f(t, z, w) dt dz dw

defines a left and right invariant Haar measure on G. Here dz is the Lebesgue measureon C = R2.

14.2. The right regular representation 1. Recall the right regular representation(R,L2(G)),

R(τ, ξ, η)f(t, z, w) = f((t, z, w)(τ, ξ, η)) = f((t+ τ, z + eitξ, w + eiαtη)) .

Let

χ(z) = e2πiRe(z) (z ∈ C) .

Then the formula

Ff(t, a, b) = f(t, a, b) =

∫C×C

χ(az + bw)f(t, z, w) dz dw

defines a bijective isometry

F : L2(G)→ L2(G) .

Lemma 117. Set

R(g) = FR(g)F−1 (g ∈ G) .

Then

F : (R,L2(G))→ (R,L2(G))

is an isomorphism of unitary representations of G and explicitly,

R(τ, ξ, η)f(t, a, b) = χ(−(ae−itξ + be−iαtη))f(t+ τ, a, b) .

80 TOMASZ PRZEBINDA

Proof. The first part is obvious. Since∫χ(az + bw)f(t+ τ, z + eitξ, w + eiαtη)) dz dw

= χ(−(ae−itξ + be−iαtη))

∫χ(az + bw)f(t, z, w) dz dw


For a, b ∈ C set

Ua,b(τ, ξ, η)v(t) = χ(−(ae−itξ + be−iαtη))v(t+ τ) (v ∈ L2(R)) .

Then a straightforward computation shows that (Ua,b,L2(R)) is a unitary representation

of G. Also, by fixing the a, b ∈ C we may view a function f(t, a, b) as a function of t andwrite

fa,b(t) = f(t, a, b) .

This way the mapL2(G) 3 f → fa,b ∈ L2(R)a,b = L2(R)

gives an isomorphism of Hilbert spaces

L2(G) =

∫C×C

L2(R)a,b da db .

Furthermore, Lemma 117 shows that the above map intertwines R with the direct integralof the representations Ua,b. Thus

(R,L2(G)) is unitarily equivalent to

∫C×C

(Ua,b,L2(R)a,b) da db . (101)

Thus


∫C×C

(Ua,b,L2(R)a,b) da db . (102)

Lemma 118. Let T = {(z1, z2) ∈ C× C; |z1| = |z2| = 1}. Then the path

P = {(eit, eiαt); t ∈ R}is not self intersecting.

Proof. Indeed, suppose(eit, eiαt) = (eis, eiαs) .

Then there are integers m, n such that

t = s+m and αt = αs+ n .

But the first equation implies that αt = αs+αm. Hence the second equations shows thatα is rational - contradiction. �

Lemma 119. For any φ ∈ C∞c (R), there is φ ∈ C∞c (T) such that

φ(t) = φ(eit, eiαt) (t ∈ R) .


Proof. This is immediate from Lemma 118. �

Lemma 120. Fix two non-zero complex numbers a and b. For any φ ∈ C∞c (R), there is

f ∈ C∞c (C× C) such that

φ(t) = f(e−ita, e−iαtb) (t ∈ R) .

Proof. This is immediate from Lemma 119. �

Lemma 121. Fix two non-zero complex numbers a and b. For any φ ∈ C∞c (R), there isf ∈ S(C× C) such that

φ(t) =

∫C×C

χ(−(ae−itξ + be−iαtη))f(ξ, η) dξ dη (t ∈ R) .

Proof. This is immediate from Lemma 120 and the fact that the Fourier transform onC× C is a bijection from S(C× C) onto itself. �

Lemma 122. Fix two non-zero complex numbers a and b. Suppose

T : L2(R)→ L2(R)

is a bounded linear map sch that

Ua,b(0, ξ, η)T = TUa,b(0, ξ, η) (ξ, η ∈ C) .

Then there is a function F such that

Tv(t) = F (t)v(t) (v ∈ L2(R) , t ∈ R) .

Proof. We see from Lemma 121 and the definition of the representation Ua,b that T com-mutes with the multiplication by functions from C∞c (R):

T (φv) = φTv (v ∈ L2(R) , φ ∈ C∞c (R)) .

Fix N > 0 and let vN ∈ C∞c (R) be equal to 1 in the interval [−N,N ]. Then for anyv ∈ C∞c (−N,N)

Tv = T (vNv) = T (vN)v .

Thus on this interval, T coincides with the multiplication by the function T (vN). Hencethere is F whose restriction to (−N,N) is equal to T (vN) and

Tv(t) = F (t)v(t) (v ∈ L2(R) , t ∈ R) .

�

Corollary 123. For any two non-zero complex numbers a and b the representation Ua,b isirreducible.

Proof. Indeed, suppose a bounded operator T : L2(R)→ L2(R) commutes with Ua,b(g) forall g ∈ G. Then Lemma 122 shows that T cincides with the multiplication by a functionF . Since

Ua,b(τ, 0, 0)v(t) = v(t+ τ)

82 TOMASZ PRZEBINDA

and since T commutes with all Ua,b(τ, 0, 0), the function F is constant. Therefore the onlyorthogonal projection that commutes with T is a constant multiple of the identity. Thisthere are no proper closed invariant subspaces in L2(R). �

Lemma 124. For a, a′, b, b′ ∈ C×, the representations Ua,b, Ua′,b′ are equivalent if and onlyif there is s ∈ R such that a′ = aeis and b′ = beiαs.

Proof. Notice that(s, 0, 0)(τ, ξ, η)(s, 0, 0)−1 = (τ, eisξ, eiαsη) .

Hence,

Ua,b(s, 0, 0)Ua,b(τ, ξ, η)Ua,b(s, 0, 0)−1 = Ua,b(τ, eisξ, eiαsη) = Uae−is,be−iαs(τ, ξ, η) .

Therefore the representations Ua,b and Uaeis,beiαs are unitarily equivalent for any s ∈ R.Conversely, if the representations Ua,b, Ua′,b′ are equivalent, then their restrictions to

the subgroup C × C ⊆ G are equivalent. But the restriction of Ua,b to C × C ⊆ G actsvia multiplication as follows

Ua,b(0, ξ, η)v(t) = χ(−(ae−itξ + be−iαtη))v(t)

Hence for any f ∈ S(C× C)

T :

∫C×C

f(ξ, η)χ(−(ae−itξ + be−iαtη))v(t) dξ dη

→∫C×C

f(ξ, η)χ(−(a′e−itξ + b′e−iαtη))Tv(t) dξ dη .

Therefore for any f ∈ S(C× C)

T : f(ae−it + be−iαt)v(t)→ f(a′e−it + b′e−iαt)Tv(t) .

Choose f ∈ C∞c ) to be concentrated very close to the path

Pa,b = {(ae−it, be−iαt); t ∈ R} .Then f(a′e−it + b′e−iαt)v(t) is zero unless f is concentrated very close to the the samepath. Which implies

{(ae−it, be−iαt); t ∈ R} = {(a′e−it, b′e−iαt); t ∈ R} .Hence there is s ∈ R such that a′ = aeis and b′ = beiαs. �

14.3. The right regular representation 2. The following formula defines a bijectionT : G→ G,

T (t, z, w) = (t, eitz, eiαtw) .

We shall use T to define a different multiplication on G,

g ◦ h = T−1(T (g)T (h)) (g, h ∈ G) .

Obviously G with the original multiplication is isomorphic to G with the new multiplica-tion. Explicitly

(t, z, w) ◦ (τ, ξ, η) = (t+ τ, e−iτz + ξ, e−iατw + η) .


It is easy to see from the above formula that the original Haar measure on G is left andright invariant with respect to the new multiplication. Let r denote the right regularrepresentation with respect to the new multiplication. Explicitly

r(τ, ξ, η)φ(t, z, w) = φ((t, z, w) ◦ (τ, ξ, η)) = φ(t+ τ, e−iτz + ξ, e−iατw + η) .

The formula

Φφ(s, a, b) = φ(s, a, b) =

∫R×C×C

χ(ts+ az + bw)φ(t, z, w) dt dz dw

defines a bijective isometryΦ : L2(G)→ L2(G) .

Lemma 125. Setr(g) = Φr(g)Φ−1 (g ∈ G) .

ThenΦ : (r,L2(G))→ (r,L2(G))

is an isomorphism of unitary representations of G and explicitly,

r(τ, ξ, η)φ(s, a, b) = χ(−(τs+ ae−iτξ + be−iατη))φ(s, e−iτa, e−iατb) .

Proof. The first part is obvious. Since∫χ(ts+ az + bw)φ((t+ τ, e−iτz + ξ, e−iατw + η) dt dz dw

= χ(−(τs+ ae−iτξ + be−iατη))

∫χ(ts+ ae−iτz + be−iατw)φ(t, z, w) dt dz dw


For r, ρ > 0 letXr,ρ = {(a, b) ∈ C× C; |a| = r , |b| = ρ} .

This is the direct product of to circles. We equip it with a measure∫Xr,ρ

ψ(a, b) dµ(a, b) =

∫Xr,ρ

ψ(reiθ1 , ρeiθ2) dθ1 dθ2 .

Then

L2(C× C) =

∫ ∞0

∫ ∞0

L2(Xr,ρ) r dr ρ dρ .

For s ∈ R and r, ρ > 0 let

Vr,ρ,s(τ, ξ, η)ψ(a, b) = χ(−(τs+ ae−iτξ + be−iατη))ψ(e−iτa, e−iατb) (ψ ∈ L2(Xr,ρ)) .

A straightforward computation shows that (Vr,ρ,s,L2(Xr,ρ)) is a unitary representation of

G (with the new multiplication).For φ ∈ S(G),∫

G

|φ(g)|2 dg =

∫R

∫ ∞0

∫ ∞0

∫Xr,ρ,s

|φ|{s}×Xr,ρ,s|2 dµ rdrdρ ds .

84 TOMASZ PRZEBINDA

Letφs,r,ρ(a, b) = φ|{s}×Xr,ρ,s(s, a, b) = ψ(s, a, b) .

Then the mapS(G) 3 φ→ φs,r,ρ ∈ L2(Xr,ρ)

intertwines r with Vr,ρ,s. Thus

(r,L2(G)) is unitarily equivalent to

∫R

∫ ∞0

∫ ∞0

(Vr,ρ,s,L2(Xr,ρ)) rdr ρdρ ds . (103)

Hence,

(r,L2(G)) is unitarily equivalent to

∫R

∫ ∞0

∫ ∞0


Thus


∫R

∫ ∞0

∫ ∞0


Problem 26. Prove that for all r, ρ and s the representations Vr,ρ,s are irreducible.

Let T be abounded operator on the Hilbert space L2(Xr,ρ) which commutes with theaction of the group. As in the (partial) proof of Theorem (116) we check that since Tcommutes with all the operators

Vr,ρ,s(0, ξ, η) (ξ, η ∈ C) ,

T coincides with the multiplication by a function F ∈ C∞(Xr,ρ). Since T commutes withall the operators

Vr,ρ,s(τ, 0, 0) (τ ∈ R) ,

the function F is constant any orbit

{(aeit, beiαt); t ∈ R} ((a, b) ∈ Xr,ρ) .

Since α is irrational any such orbit is dense in Xr,ρ. Hence F is constant. Thus T coincideswith a constant multiple of the identity.

14.4. A surprise.None of the representations (Ua,b,L

2(R)) is equivalent to any (Vr,ρ,s,L2(Xr,ρ)). Indeed,

consider the restriction of both to the abelian subgroup

N = {(0, z, w); z, w ∈ C} .Thus

Ua,b(0, ξ, η)v(t) = χ(−(e−itaξ + e−iαtbη))v(t)

andVr,ρ,s(0, ξ, η)ψ(a, b) = χ(−(aξ + bη))ψ(a, b) .

If the two representation would be equivalent then the two characters

C× C 3 (ξ, η)→ χ(−(e−itaξ + e−iαtbη)) ∈ C× ,C× C 3 (ξ, η)→ χ(−(aξ + bη)) ∈ C×


would be equal. But they are not (for almost all t)!

References

[Bar47] V. Bargmann. Irreducible unitary representations of the lorentz group. Annals of Math., 48:568–640, 1947.

[Bar89] D. Barbasch. The unitary dual for complex classical Lie groups. Invent. Math., 96:103–176, 1989.[FH91] W. Fulton and J. Harris. Representation Theory, a first course. Springer Verlag, 1991.[Har76] Harish-Chandra. Harmonic Analysis on Real Reductive Groups III, The Mass - Selberg relations

and the Plancherel formula. Ann. of Math., 104:117–201, 1976.[HC52] Harish-Chandra. Plancherel formula for the 2× 2 real unimodular group. Proc. Nat. Acad. Sci.

U. S. A., 38:337–342, 1952.[HC14a] Harish-Chandra. Collected papers. I. 1944–1954. Springer Collected Works in Mathematics.

Springer, New York, 2014. Edited and with an introduction by V. S. Varadarajan, With intro-ductory essays by Nolan R. Wallach and Roger Howe, Reprint of the 1984 edition [ MR0726025].

[HC14b] Harish-Chandra. Collected papers. II. 1955–1958. Springer Collected Works in Mathematics.Springer, Heidelberg, 2014. Edited and with an introduction by V. S. Varadarajan, With intro-ductory essays by Nolan R. Wallach and Roger Howe, Reprint of the 1984 edition [ MR0726023].

[HC14c] Harish-Chandra. Collected papers. III. 1959–1968. Springer Collected Works in Mathematics.Springer, Heidelberg, 2014. Edited and with an introduction by V. S. Varadarajan, With intro-ductory essays by Nolan R. Wallach and Roger Howe, Reprint of the 1984 edition [ MR0726024].

[HC14d] Harish-Chandra. Collected papers. IV. 1970–1983. Springer Collected Works in Mathematics.Springer, Heidelberg, 2014. Edited and with an introduction by V. S. Varadarajan, With intro-ductory essays by Nolan R. Wallach and Roger Howe, Reprint of the 1984 edition [ MR0726026].

[HC18] Harish-Chandra. Collected papers. V (Posthumous). Harmonic analysis in real semisimplegroups. Springer, Cham, 2018. Edited by Ramesh Gangolli and V. S. Varadarajan, with as-sistance from Johan Kolk.

[Hor83] L. Hormander. The Analysis of Linear Partial Differential Operators I. Springer Verlag, 1983.[HR63] E. Hewitt and K. A. Ross. Abstract Harmonic Analysis. Springer Verlag, 1963.[HT92] R. Howe and E. Tan. Nonabelian harmonic analysis. Applications of SL(2, R). Springer Verlag,

1992.[Kir78] A. A. Kirillov. Elements of the theory of representations. Nauka, Moscow, 1978.[Kna86] A. Knapp. Representation Theory of Semisimple groups, an overview based on examples. Prince-

ton Mathematical Series. Princeton University Press, Princeton, New Jersey, 1986.[Lan75] Serge Lang. SL2(R). Addison-Wesley Publishing Co., Reading, Mass.-London-Amsterdam, 1975.[Lan85] S. Lang. SL2(R). Springer-Verlag, 1985.[Par06] M.-A. Parseval. Memoire sur les series et sur l’intgration complete d’une equation

aux differences partielles lineaire du second ordre, a coefficients constants.Memoires presentes a l’Institut des Sciences, Lettres et Arts, par divers savants, et lus dans ses assemblees. Sciences, mathematiques et physiques. (Savants etrangers.),1:638–648, 1806.

[Pla10] M. Plancherel. Contribution A ’Etude de la reprEsentation D’une fonction arbitraire par

des intEgrales dEfinies. Rendiconti del Circolo Matematico di Palermo (1884-1940), 30:289–335,1910.

[PW27] F. Peter and H. Weyl. Die Vollstandigkeit der primitiven Darstellungen einer geschlossenenkontinuierlichen Gruppe. Math. Ann., 97(1):737–755, 1927.

[Rud91] Rudin, W. Functional Analysis. McGraw-Hill, Inc, 1991.[SS03] E.M. Stein and R. Shakarchi. Fourier Analysis: An Introduction (Princetone Lectures in

Analysis). Princeton University Press, Princeton, NJ, 2003.[SvN50] I. Segal and J. von Neumann. A theorem on unitary representations of semisimple Lie groups.

Ann. of Math., 52:509–517, 1950.

86 TOMASZ PRZEBINDA

[Var89] V. S. Varadarajan. An introduction to harmonic analysis semisimple groups. Cambridge Uni-versity press, 1989.

[vN26] John von Neumann. Die Eindeutigkeit der schrodingerschen Operatoren. Math. Ann., 96:737–755, 1926.

[Wal88] N. Wallach. Real Reductive Groups I. Academic Press, 1988.

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Contentstomasz.przebinda.com/LieTheoryII_Spring_2020.pdf · Problems arising in Physics and Number Theory motivated a rapid growth of Harmonic Analysis on non-commutative groups